{"id":543385,"date":"2021-09-30T04:19:40","date_gmt":"2021-09-30T04:19:40","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=543385"},"modified":"2021-10-01T06:45:00","modified_gmt":"2021-10-01T06:45:00","slug":"rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/","title":{"rendered":"RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations"},"content":{"rendered":"\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 11: Maths Chapter 11 solutions. Complete Class 11 Maths Chapter 11 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\">RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations<\/h2>\n\n\n\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">RD Sharma 11th Maths Chapter 11, Class 11 Maths Chapter 11 solutions<\/p>\n\n\n\n<p><strong>1. Find the general solutions of the following equations:<\/strong><\/p>\n\n\n\n<p><strong>(i) sin x = 1\/2<\/strong><\/p>\n\n\n\n<p><strong>(ii) cos x = \u2013 \u221a3\/2<\/strong><\/p>\n\n\n\n<p><strong>(iii) cosec x = \u2013 \u221a2<\/strong><\/p>\n\n\n\n<p><strong>(iv) sec x = \u221a2<\/strong><\/p>\n\n\n\n<p><strong>(v) tan x = -1\/\u221a3<\/strong><\/p>\n\n\n\n<p><strong>(vi) \u221a3 sec x = 2<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The general solution of any trigonometric equation is given as:<\/p>\n\n\n\n<p>sin x = sin y, implies x = n\u03c0 + (\u2013 1)<sup>n&nbsp;<\/sup>y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>cos x = cos y, implies x = 2n\u03c0&nbsp;\u00b1&nbsp;y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>tan x = tan y, implies x = n\u03c0&nbsp;+ y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin x = 1\/2<\/p>\n\n\n\n<p>We know sin 30<sup>o<\/sup>&nbsp;= sin \u03c0\/6 = \u00bd<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>Sin x = sin \u03c0\/6<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 + (\u2013 1)<sup>&nbsp;n&nbsp;<\/sup>\u03c0\/6, where n&nbsp;\u2208&nbsp;Z. [since, sin x = sin A =&gt; x = n\u03c0 + (\u2013 1)<sup>&nbsp;n<\/sup>&nbsp;A]<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>cos x = \u2013 \u221a3\/2<\/p>\n\n\n\n<p>We know, cos 150<sup>o<\/sup>&nbsp;= (- \u221a3\/2) = cos 5\u03c0\/6<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>Cos x = cos 5\u03c0\/6<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = 2n\u03c0&nbsp;\u00b1 5\u03c0\/6, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>cosec x = \u2013 \u221a2<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>1\/sin x = \u2013 \u221a2 [since, cosec x = 1\/sin x]<\/p>\n\n\n\n<p>Sin x = -1\/\u221a2<\/p>\n\n\n\n<p>= sin [\u03c0 + \u03c0\/4]<\/p>\n\n\n\n<p>= sin 5\u03c0\/4 or sin (-\u03c0\/4)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 + (-1)<sup>n+1<\/sup>&nbsp;\u03c0\/4, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iv)&nbsp;<\/strong>sec x = \u221a2<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>1\/cos x = \u221a2 [since, sec x = 1\/cos x]<\/p>\n\n\n\n<p>Cos x = 1\/\u221a2<\/p>\n\n\n\n<p>= cos \u03c0\/4<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = 2n\u03c0&nbsp;\u00b1 \u03c0\/4, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>tan x = -1\/\u221a3<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan x = -1\/\u221a3<\/p>\n\n\n\n<p>tan x = tan (\u03c0\/6)<\/p>\n\n\n\n<p>= tan (-\u03c0\/6) [since, tan (-x) = -tan x]<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 + (-\u03c0\/6), where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p>or x = n\u03c0 \u2013 \u03c0\/6, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(vi)&nbsp;<\/strong>\u221a3 sec x = 2<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sec x = 2\/\u221a3<\/p>\n\n\n\n<p>1\/cos x = 2\/\u221a3<\/p>\n\n\n\n<p>Cos x = \u221a3\/2<\/p>\n\n\n\n<p>= cos (\u03c0\/6)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = 2n\u03c0&nbsp;\u00b1 \u03c0\/6, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>2. Find the general solutions of the following equations:<\/strong><\/p>\n\n\n\n<p><strong>(i) sin 2x = \u221a3\/2<\/strong><\/p>\n\n\n\n<p><strong>(ii) cos 3x = 1\/2<\/strong><\/p>\n\n\n\n<p><strong>(iii) sin 9x = sin x<\/strong><\/p>\n\n\n\n<p><strong>(iv) sin 2x = cos 3x<\/strong><\/p>\n\n\n\n<p><strong>(v) tan x + cot 2x = 0<\/strong><\/p>\n\n\n\n<p><strong>(vi) tan 3x = cot x<\/strong><\/p>\n\n\n\n<p><strong>(vii) tan 2x tan x = 1<\/strong><\/p>\n\n\n\n<p><strong>(viii) tan mx + cot nx = 0<\/strong><\/p>\n\n\n\n<p><strong>(ix) tan px = cot qx<\/strong><\/p>\n\n\n\n<p><strong>(x) sin 2x + cos x = 0<\/strong><\/p>\n\n\n\n<p><strong>(xi) sin x = tan x<\/strong><\/p>\n\n\n\n<p><strong>(xii) sin 3x + cos 2x = 0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The general solution of any trigonometric equation is given as:<\/p>\n\n\n\n<p>sin x = sin y, implies x = n\u03c0 + (\u2013 1)<sup>n&nbsp;<\/sup>y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>cos x = cos y, implies x = 2n\u03c0&nbsp;\u00b1&nbsp;y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>tan x = tan y, implies x = n\u03c0&nbsp;+ y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin 2x = \u221a3\/2<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin 2x = \u221a3\/2<\/p>\n\n\n\n<p>= sin (\u03c0\/3)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>2x = n\u03c0 + (-1)<sup>n<\/sup>&nbsp;\u03c0\/3, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p>x = n\u03c0\/2 + (-1)<sup>n<\/sup>&nbsp;\u03c0\/6, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>cos 3x = 1\/2<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>cos 3x = 1\/2<\/p>\n\n\n\n<p>= cos (\u03c0\/3)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>3x = 2n\u03c0 \u00b1 \u03c0\/3, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p>x = 2n\u03c0\/3 \u00b1 \u03c0\/9, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>sin 9x = sin x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>Sin 9x \u2013 sin x = 0<\/p>\n\n\n\n<p>Using transformation formula,<\/p>\n\n\n\n<p>Sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>= 2 cos (9x+x)\/2 sin (9x-x)\/2<\/p>\n\n\n\n<p>=&gt; cos 5x sin 4x = 0<\/p>\n\n\n\n<p>Cos 5x = 0 or sin 4x = 0<\/p>\n\n\n\n<p>Let us verify both the expressions,<\/p>\n\n\n\n<p>Cos 5x = 0<\/p>\n\n\n\n<p>Cos 5x = cos \u03c0\/2<\/p>\n\n\n\n<p>5x = (2n + 1)\u03c0\/2<\/p>\n\n\n\n<p>x = (2n + 1)\u03c0\/10, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p>sin 4x = 0<\/p>\n\n\n\n<p>sin 4x = sin 0<\/p>\n\n\n\n<p>4x = n\u03c0<\/p>\n\n\n\n<p>x = n\u03c0\/4, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = (2n + 1)\u03c0\/10 or n\u03c0\/4, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iv)&nbsp;<\/strong>sin 2x = cos 3x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin 2x = cos 3x<\/p>\n\n\n\n<p>cos (\u03c0\/2 \u2013 2x) = cos 3x [since, sin A = cos (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>\u03c0\/2 \u2013 2x = 2n\u03c0 \u00b1 3x<\/p>\n\n\n\n<p>\u03c0\/2 \u2013 2x = 2n\u03c0 + 3x [or] \u03c0\/2 \u2013 2x = 2n\u03c0 \u2013 3x<\/p>\n\n\n\n<p>5x = \u03c0\/2 + 2n\u03c0 [or] x = 2n\u03c0 \u2013 \u03c0\/2<\/p>\n\n\n\n<p>5x = \u03c0\/2 (1 + 4n) [or] x = \u03c0\/2 (4n \u2013 1)<\/p>\n\n\n\n<p>x = \u03c0\/10 (1 + 4n) [or] x = \u03c0\/2 (4n \u2013 1)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = \u03c0\/10 (4n + 1) [or] x = \u03c0\/2 (4n \u2013 1), where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>tan x + cot 2x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan x = \u2013 cot 2x<\/p>\n\n\n\n<p>tan x = \u2013 tan (\u03c0\/2 \u2013 2x) [since, cot A = tan (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>tan x = tan (2x \u2013 \u03c0\/2) [since, \u2013 tan A = tan -A]<\/p>\n\n\n\n<p>x = n\u03c0 + 2x \u2013 \u03c0\/2<\/p>\n\n\n\n<p>x = n\u03c0 \u2013 \u03c0\/2<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 \u2013 \u03c0\/2, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(vi)&nbsp;<\/strong>tan 3x = cot x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan 3x = cot x<\/p>\n\n\n\n<p>tan 3x = tan (\u03c0\/2 \u2013 x) [since, cot A = tan (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>3x = n\u03c0 + \u03c0\/2 \u2013 x<\/p>\n\n\n\n<p>4x = n\u03c0 + \u03c0\/2<\/p>\n\n\n\n<p>x = n\u03c0\/4 + \u03c0\/8<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0\/4 + \u03c0\/8, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(vii)&nbsp;<\/strong>tan 2x tan x = 1<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan 2x tan x = 1<\/p>\n\n\n\n<p>tan 2x = 1\/tan x<\/p>\n\n\n\n<p>= cot x<\/p>\n\n\n\n<p>tan 2x = tan (\u03c0\/2 \u2013 x) [since, cot A = tan (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>2x = n\u03c0 + \u03c0\/2 \u2013 x<\/p>\n\n\n\n<p>3x = n\u03c0 + \u03c0\/2<\/p>\n\n\n\n<p>x = n\u03c0\/3 + \u03c0\/6<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0\/3 + \u03c0\/6, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(viii)&nbsp;<\/strong>tan mx + cot nx = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan mx + cot nx = 0<\/p>\n\n\n\n<p>tan mx = \u2013 cot nx<\/p>\n\n\n\n<p>= \u2013 tan (\u03c0\/2 \u2013 nx) [since, cot A = tan (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>tan mx = tan (nx + \u03c0\/2) [since, \u2013 tan A = tan -A]<\/p>\n\n\n\n<p>mx = k\u03c0 + nx + \u03c0\/2<\/p>\n\n\n\n<p>(m \u2013 n) x = k\u03c0 + \u03c0\/2<\/p>\n\n\n\n<p>(m \u2013 n) x = \u03c0 (2k + 1)\/2<\/p>\n\n\n\n<p>x = \u03c0 (2k + 1)\/2(m \u2013 n)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = \u03c0 (2k + 1)\/2(m \u2013 n), where m, n, k&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(ix)&nbsp;<\/strong>tan px = cot qx<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan px = cot qx<\/p>\n\n\n\n<p>tan px = tan (\u03c0\/2 \u2013 qx) [since, cot A = tan (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>px = n\u03c0 \u00b1 (\u03c0\/2 \u2013 qx)<\/p>\n\n\n\n<p>(p + q) x = n\u03c0 + \u03c0\/2<\/p>\n\n\n\n<p>x = n\u03c0\/(p+q) + \u03c0\/2(p+q)<\/p>\n\n\n\n<p>= \u03c0 (2n +1)\/ 2(p+q)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = \u03c0 (2n +1)\/ 2(p+q), where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(x)&nbsp;<\/strong>sin 2x + cos x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin 2x + cos x = 0<\/p>\n\n\n\n<p>cos x = \u2013 sin 2x<\/p>\n\n\n\n<p>cos x = \u2013 cos (\u03c0\/2 \u2013 2x) [since, sin A = cos (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>= cos (\u03c0 \u2013 (\u03c0\/2 \u2013 2x)) [since, -cos A = cos (\u03c0 \u2013 A)]<\/p>\n\n\n\n<p>= cos (\u03c0\/2 + 2x)<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 (\u03c0\/2 + 2x)<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>x = 2n\u03c0 + (\u03c0\/2 + 2x) [or] x = 2n\u03c0 \u2013 (\u03c0\/2 + 2x)<\/p>\n\n\n\n<p>x = \u2013 \u03c0\/2 \u2013 2n\u03c0 [or] 3x = 2n\u03c0 \u2013 \u03c0\/2<\/p>\n\n\n\n<p>x = \u2013 \u03c0\/2 (1 + 4n) [or] x = \u03c0\/6 (4n \u2013 1)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = \u2013 \u03c0\/2 (1 + 4n), where n&nbsp;\u03f5&nbsp;Z. [or] x = \u03c0\/6 (4n \u2013 1)<\/p>\n\n\n\n<p>x = \u03c0\/2 (4n \u2013 1), where n&nbsp;\u03f5&nbsp;Z. [or] x = \u03c0\/6 (4n \u2013 1), where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(xi)&nbsp;<\/strong>sin x = tan x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin x = tan x<\/p>\n\n\n\n<p>sin x = sin x\/cos x<\/p>\n\n\n\n<p>sin x cos x = sin x<\/p>\n\n\n\n<p>sin x (cos x \u2013 1) = 0<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>Sin x = 0 or cos x \u2013 1 = 0<\/p>\n\n\n\n<p>Sin x = sin 0 [or] cos x = 1<\/p>\n\n\n\n<p>Sin x = sin 0 [or] cos x = cos 0<\/p>\n\n\n\n<p>x = n\u03c0 [or] x = 2m\u03c0<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 [or] 2m\u03c0, where n, m&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(xii)&nbsp;<\/strong>sin 3x + cos 2x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin 3x + cos 2x = 0<\/p>\n\n\n\n<p>cos 2x = \u2013 sin 3x<\/p>\n\n\n\n<p>cos 2x = \u2013 cos (\u03c0\/2 \u2013 3x) [since, sin A = cos (\u03c0\/2 \u2013 A)]<\/p>\n\n\n\n<p>cos 2x = cos (\u03c0 \u2013 (\u03c0\/2 \u2013 3x)) [since, -cos A = cos (\u03c0 \u2013 A)]<\/p>\n\n\n\n<p>cos 2x = cos (\u03c0\/2 + 3x)<\/p>\n\n\n\n<p>2x = 2n\u03c0 \u00b1 (\u03c0\/2 + 3x)<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>2x = 2n\u03c0 + (\u03c0\/2 + 3x) [or] 2x = 2n\u03c0 \u2013 (\u03c0\/2 + 3x)<\/p>\n\n\n\n<p>x = -\u03c0\/2 \u2013 2n\u03c0 [or] 5x = 2n\u03c0 \u2013 \u03c0\/2<\/p>\n\n\n\n<p>x = -\u03c0\/2 (1 + 4n) [or] x = \u03c0\/10 (4n \u2013 1)<\/p>\n\n\n\n<p>x = \u2013 \u03c0\/2 (4n + 1) [or] \u03c0\/10 (4n \u2013 1)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = \u2013 \u03c0\/2 (4n + 1) [or] \u03c0\/10 (4n \u2013 1)<\/p>\n\n\n\n<p>x = \u03c0\/2 (4n \u2013 1) [or] \u03c0\/10 (4n \u2013 1), where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>3. Solve the following equations:<\/strong><\/p>\n\n\n\n<p><strong>(i) sin<sup>2<\/sup>&nbsp;x \u2013 cos x = 1\/4<\/strong><\/p>\n\n\n\n<p><strong>(ii) 2 cos<sup>2<\/sup>&nbsp;x \u2013 5 cos x + 2 = 0<\/strong><\/p>\n\n\n\n<p><strong>(iii) 2 sin<sup>2<\/sup>&nbsp;x + \u221a3 cos x + 1 = 0<\/strong><\/p>\n\n\n\n<p><strong>(iv) 4 sin<sup>2<\/sup>&nbsp;x \u2013 8 cos x + 1 = 0<\/strong><\/p>\n\n\n\n<p><strong>(v) tan<sup>2<\/sup>&nbsp;x + (1 \u2013 \u221a3) tan x \u2013 \u221a3 = 0<\/strong><\/p>\n\n\n\n<p><strong>(vi) 3 cos<sup>2<\/sup>&nbsp;x \u2013 2\u221a3 sin x cos x \u2013 3 sin<sup>2<\/sup>&nbsp;x = 0<\/strong><\/p>\n\n\n\n<p><strong>(vii) cos 4x = cos 2x<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The general solution of any trigonometric equation is given as:<\/p>\n\n\n\n<p>sin x = sin y, implies x = n\u03c0 + (\u2013 1)<sup>n&nbsp;<\/sup>y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>cos x = cos y, implies x = 2n\u03c0&nbsp;\u00b1&nbsp;y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>tan x = tan y, implies x = n\u03c0&nbsp;+ y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>sin<sup>2<\/sup>&nbsp;x \u2013 cos x = 1\/4<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin<sup>2<\/sup>&nbsp;x \u2013 cos x = \u00bc<\/p>\n\n\n\n<p>1 \u2013 cos<sup>2<\/sup>&nbsp;x \u2013 cos x = 1\/4 [since, sin<sup>2<\/sup>&nbsp;x = 1 \u2013 cos<sup>2<\/sup>&nbsp;x]<\/p>\n\n\n\n<p>4 \u2013 4 cos<sup>2<\/sup>&nbsp;x \u2013 4 cos x = 1<\/p>\n\n\n\n<p>4cos<sup>2<\/sup>&nbsp;x + 4cos x \u2013 3 = 0<\/p>\n\n\n\n<p>Let cos x be \u2018k\u2019<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>4k<sup>2<\/sup>&nbsp;+ 4k \u2013 3 = 0<\/p>\n\n\n\n<p>4k<sup>2<\/sup>&nbsp;\u2013 2k + 6k \u2013 3 = 0<\/p>\n\n\n\n<p>2k (2k \u2013 1) + 3 (2k \u2013 1) = 0<\/p>\n\n\n\n<p>(2k \u2013 1) + (2k + 3) = 0<\/p>\n\n\n\n<p>(2k \u2013 1) = 0 or (2k + 3) = 0<\/p>\n\n\n\n<p>k = 1\/2 or k = -3\/2<\/p>\n\n\n\n<p>cos x = 1\/2 or cos x = -3\/2<\/p>\n\n\n\n<p>we shall consider only cos x = 1\/2. cos x = -3\/2 is not possible.<\/p>\n\n\n\n<p>so,<\/p>\n\n\n\n<p>cos x = cos 60<sup>o<\/sup>&nbsp;= cos \u03c0\/3<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 \u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 \u03c0\/3, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>2 cos<sup>2<\/sup>&nbsp;x \u2013 5 cos x + 2 = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>2 cos<sup>2<\/sup>&nbsp;x \u2013 5 cos x + 2 = 0<\/p>\n\n\n\n<p>Let cos x be \u2018k\u2019<\/p>\n\n\n\n<p>2k<sup>2<\/sup>&nbsp;\u2013 5k + 2 = 0<\/p>\n\n\n\n<p>2k<sup>2<\/sup>&nbsp;\u2013 4k \u2013 k +2 = 0<\/p>\n\n\n\n<p>2k(k \u2013 2) -1(k -2) = 0<\/p>\n\n\n\n<p>(k \u2013 2) (2k \u2013 1) = 0<\/p>\n\n\n\n<p>k = 2 or k = 1\/2<\/p>\n\n\n\n<p>cos x = 2 or cos x = 1\/2<\/p>\n\n\n\n<p>we shall consider only cos x = 1\/2. cos x = 2 is not possible.<\/p>\n\n\n\n<p>so,<\/p>\n\n\n\n<p>cos x = cos 60<sup>o<\/sup>&nbsp;= cos \u03c0\/3<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 \u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 \u03c0\/3, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>2 sin<sup>2<\/sup>&nbsp;x + \u221a3 cos x + 1 = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>2 sin<sup>2<\/sup>&nbsp;x + \u221a3 cos x + 1 = 0<\/p>\n\n\n\n<p>2 (1 \u2013 cos<sup>2<\/sup>&nbsp;x) + \u221a3 cos x + 1 = 0 [since, sin<sup>2<\/sup>&nbsp;x = 1 \u2013 cos<sup>2<\/sup>&nbsp;x]<\/p>\n\n\n\n<p>2 \u2013 2 cos<sup>2<\/sup>&nbsp;x + \u221a3 cos x + 1 = 0<\/p>\n\n\n\n<p>2 cos<sup>2<\/sup>&nbsp;x \u2013 \u221a3 cos x \u2013 3 = 0<\/p>\n\n\n\n<p>Let cos x be \u2018k\u2019<\/p>\n\n\n\n<p>2k<sup>2<\/sup>&nbsp;\u2013 \u221a3 k \u2013 3 = 0<\/p>\n\n\n\n<p>2k<sup>2<\/sup>&nbsp;-2\u221a3 k + \u221a3 k \u2013 3 = 0<\/p>\n\n\n\n<p>2k(k \u2013 \u221a3) +\u221a3(k \u2013 \u221a3) = 0<\/p>\n\n\n\n<p>(2k + \u221a3) (k \u2013 \u221a3) = 0<\/p>\n\n\n\n<p>k = \u221a3 or k = -\u221a3\/2<\/p>\n\n\n\n<p>cos x = \u221a3 or cos x = -\u221a3\/2<\/p>\n\n\n\n<p>we shall consider only cos x = -\u221a3\/2. cos x = \u221a3 is not possible.<\/p>\n\n\n\n<p>so,<\/p>\n\n\n\n<p>cos x = -\u221a3\/2<\/p>\n\n\n\n<p>cos x = cos 150\u00b0 = cos 5\u03c0\/6<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 5\u03c0\/6, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iv)&nbsp;<\/strong>4 sin<sup>2<\/sup>&nbsp;x \u2013 8 cos x + 1 = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>4 sin<sup>2<\/sup>&nbsp;x \u2013 8 cos x + 1 = 0<\/p>\n\n\n\n<p>4 (1 \u2013 cos<sup>2<\/sup>&nbsp;x) \u2013 8 cos x + 1 = 0 [since, sin<sup>2<\/sup>&nbsp;x = 1 \u2013 cos<sup>2<\/sup>&nbsp;x]<\/p>\n\n\n\n<p>4 \u2013 4 cos<sup>2<\/sup>&nbsp;x \u2013 8 cos x + 1 = 0<\/p>\n\n\n\n<p>4 cos<sup>2<\/sup>&nbsp;x + 8 cos x \u2013 5 = 0<\/p>\n\n\n\n<p>Let cos x be \u2018k\u2019<\/p>\n\n\n\n<p>4k<sup>2<\/sup>&nbsp;+ 8k \u2013 5 = 0<\/p>\n\n\n\n<p>4k<sup>2<\/sup>&nbsp;-2k + 10k \u2013 5 = 0<\/p>\n\n\n\n<p>2k(2k \u2013 1) + 5(2k \u2013 1) = 0<\/p>\n\n\n\n<p>(2k + 5) (2k \u2013 1) = 0<\/p>\n\n\n\n<p>k = -5\/2 = -2.5 or k = 1\/2<\/p>\n\n\n\n<p>cos x = -2.5 or cos x = 1\/2<\/p>\n\n\n\n<p>we shall consider only cos x = 1\/2. cos x = -2.5 is not possible.<\/p>\n\n\n\n<p>so,<\/p>\n\n\n\n<p>cos x = cos 60<sup>o<\/sup>&nbsp;= cos \u03c0\/3<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 \u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = 2n\u03c0 \u00b1 \u03c0\/3, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>tan<sup>2<\/sup>&nbsp;x + (1 \u2013 \u221a3) tan x \u2013 \u221a3 = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan<sup>2<\/sup>&nbsp;x + (1 \u2013 \u221a3) tan x \u2013 \u221a3 = 0<\/p>\n\n\n\n<p>tan<sup>2<\/sup>&nbsp;x + tan x \u2013 \u221a3 tan x \u2013 \u221a3 = 0<\/p>\n\n\n\n<p>tan x (tan x + 1) \u2013 \u221a3 (tan x + 1) = 0<\/p>\n\n\n\n<p>(tan x + 1) ( tan x \u2013 \u221a3) = 0<\/p>\n\n\n\n<p>tan x = -1 or tan x = \u221a3<\/p>\n\n\n\n<p>As, tan x&nbsp;\u03f5&nbsp;(-\u221e , \u221e) so both values are valid and acceptable.<\/p>\n\n\n\n<p>tan x = tan (-\u03c0\/4) or tan x = tan (\u03c0\/3)<\/p>\n\n\n\n<p>x = m\u03c0 \u2013 \u03c0\/4 or x = n\u03c0 + \u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = m\u03c0 \u2013 \u03c0\/4 or n\u03c0 + \u03c0\/3, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(vi)&nbsp;<\/strong>3 cos<sup>2<\/sup>&nbsp;x \u2013 2\u221a3 sin x cos x \u2013 3 sin<sup>2<\/sup>&nbsp;x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>3 cos<sup>2<\/sup>&nbsp;x \u2013 2\u221a3 sin x cos x \u2013 3 sin<sup>2<\/sup>&nbsp;x = 0<\/p>\n\n\n\n<p>3 cos<sup>2<\/sup>&nbsp;x \u2013 3\u221a3 sin x cos x + \u221a3 sin x cos x \u2013 3 sin<sup>2<\/sup>&nbsp;x = 0<\/p>\n\n\n\n<p>3 cos x (cos x \u2013 \u221a3sin x) + \u221a3 sin x (cos x \u2013 \u221a3 sin x) = 0<\/p>\n\n\n\n<p>\u221a3 (cos x \u2013 \u221a3 sin x) (\u221a3 cos x + sin x) = 0<\/p>\n\n\n\n<p>cos x \u2013 \u221a3 sin x = 0 or sin x + \u221a3 cos x = 0<\/p>\n\n\n\n<p>cos x = \u221a3 sin x or sin x = -\u221a3 cos x<\/p>\n\n\n\n<p>tan x = 1\/\u221a3 or tan x = -\u221a3<\/p>\n\n\n\n<p>As, tan x&nbsp;\u03f5&nbsp;(-\u221e , \u221e) so both values are valid and acceptable.<\/p>\n\n\n\n<p>tan x = tan (\u03c0\/6) or tan x = tan (-\u03c0\/3)<\/p>\n\n\n\n<p>x = m\u03c0 + \u03c0\/6 or x = n\u03c0 \u2013 \u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = m\u03c0 + \u03c0\/6 or n\u03c0 \u2013 \u03c0\/3, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(vii)&nbsp;<\/strong>cos 4x = cos 2x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>cos 4x = cos 2x<\/p>\n\n\n\n<p>4x = 2n\u03c0 \u00b1 2x<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>4x = 2n\u03c0 + 2x [or] 4x = 2n\u03c0 \u2013 2x<\/p>\n\n\n\n<p>2x = 2n\u03c0 [or] 6x = 2n\u03c0<\/p>\n\n\n\n<p>x = n\u03c0 [or] x = n\u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 [or] n\u03c0\/3, where n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>4. Solve the following equations:<\/strong><\/p>\n\n\n\n<p><strong>(i) cos x + cos 2x + cos 3x = 0<\/strong><\/p>\n\n\n\n<p><strong>(ii) cos x + cos 3x \u2013 cos 2x = 0<\/strong><\/p>\n\n\n\n<p><strong>(iii) sin x + sin 5x = sin 3x<\/strong><\/p>\n\n\n\n<p><strong>(iv) cos x cos 2x cos 3x = 1\/4<\/strong><\/p>\n\n\n\n<p><strong>(v) cos x + sin x = cos 2x + sin 2x<\/strong><\/p>\n\n\n\n<p><strong>(vi) sin x + sin 2x + sin 3x = 0<\/strong><\/p>\n\n\n\n<p><strong>(vii) sin x + sin 2x + sin 3x + sin 4x = 0<\/strong><\/p>\n\n\n\n<p><strong>(viii) sin 3x \u2013 sin x = 4 cos<sup>2<\/sup>&nbsp;x \u2013 2<\/strong><\/p>\n\n\n\n<p><strong>(ix) sin 2x \u2013 sin 4x + sin 6x = 0<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The general solution of any trigonometric equation is given as:<\/p>\n\n\n\n<p>sin x = sin y, implies x = n\u03c0 + (\u2013 1)<sup>n&nbsp;<\/sup>y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>cos x = cos y, implies x = 2n\u03c0&nbsp;\u00b1&nbsp;y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>tan x = tan y, implies x = n\u03c0&nbsp;+ y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>cos x + cos 2x + cos 3x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>cos x + cos 2x + cos 3x = 0<\/p>\n\n\n\n<p>we shall rearrange and use transformation formula<\/p>\n\n\n\n<p>cos 2x + (cos x + cos 3x) = 0<\/p>\n\n\n\n<p>by using the formula, cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>cos 2x + 2 cos (3x+x)\/2 cos (3x-x)\/2 = 0<\/p>\n\n\n\n<p>cos 2x + 2cos 2x cos x = 0<\/p>\n\n\n\n<p>cos 2x ( 1 + 2 cos x) = 0<\/p>\n\n\n\n<p>cos 2x = 0 or 1 + 2cos x = 0<\/p>\n\n\n\n<p>cos 2x = cos 0 or cos x = -1\/2<\/p>\n\n\n\n<p>cos 2x = cos \u03c0\/2 or cos x = cos (\u03c0 \u2013 \u03c0\/3)<\/p>\n\n\n\n<p>cos 2x = cos \u03c0\/2 or cos x = cos (2\u03c0\/3)<\/p>\n\n\n\n<p>2x = (2n + 1) \u03c0\/2 or x = 2m\u03c0 \u00b1 2\u03c0\/3<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/4 or x = 2m\u03c0 \u00b1 2\u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/4 or 2m\u03c0 \u00b1 2\u03c0\/3, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>cos x + cos 3x \u2013 cos 2x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>cos x + cos 3x \u2013 cos 2x = 0<\/p>\n\n\n\n<p>we shall rearrange and use transformation formula<\/p>\n\n\n\n<p>cos x \u2013 cos 2x + cos 3x = 0<\/p>\n\n\n\n<p>\u2013 cos 2x + (cos x + cos 3x) = 0<\/p>\n\n\n\n<p>By using the formula, cos A + cos B = 2 cos (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>\u2013 cos 2x + 2 cos (3x+x)\/2 cos (3x-x)\/2 = 0<\/p>\n\n\n\n<p>\u2013 cos 2x + 2cos 2x cos x = 0<\/p>\n\n\n\n<p>cos 2x ( -1 + 2 cos x) = 0<\/p>\n\n\n\n<p>cos 2x = 0 or -1 + 2cos x = 0<\/p>\n\n\n\n<p>cos 2x = cos 0 or cos x = 1\/2<\/p>\n\n\n\n<p>cos 2x = cos \u03c0\/2 or cos x = cos (\u03c0\/3)<\/p>\n\n\n\n<p>2x = (2n + 1) \u03c0\/2 or x = 2m\u03c0 \u00b1 \u03c0\/3<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/4 or x = 2m\u03c0 \u00b1 \u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/4 or 2m\u03c0 \u00b1 \u03c0\/3, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>sin x + sin 5x = sin 3x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin x + sin 5x = sin 3x<\/p>\n\n\n\n<p>sin x + sin 5x \u2013 sin 3x = 0<\/p>\n\n\n\n<p>we shall rearrange and use transformation formula<\/p>\n\n\n\n<p>\u2013 sin 3x + sin x + sin 5x = 0<\/p>\n\n\n\n<p>\u2013 sin 3x + (sin 5x + sin x) = 0<\/p>\n\n\n\n<p>By using the formula, sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>\u2013 sin 3x + 2 sin (5x+x)\/2 cos (5x-x)\/2 = 0<\/p>\n\n\n\n<p>2sin 3x cos 2x \u2013 sin 3x = 0<\/p>\n\n\n\n<p>sin 3x ( 2cos 2x \u2013 1) = 0<\/p>\n\n\n\n<p>sin 3x = 0 or 2cos 2x \u2013 1 = 0<\/p>\n\n\n\n<p>sin 3x = sin 0 or cos 2x = 1\/2<\/p>\n\n\n\n<p>sin 3x = sin 0 or cos 2x = cos \u03c0\/3<\/p>\n\n\n\n<p>3x = n\u03c0 or 2x = 2m\u03c0 \u00b1 \u03c0\/3<\/p>\n\n\n\n<p>x = n\u03c0\/3 or x = m\u03c0 \u00b1 \u03c0\/6<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0\/3 or m\u03c0 \u00b1 \u03c0\/6, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iv)&nbsp;<\/strong>cos x cos 2x cos 3x = 1\/4<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>cos x cos 2x cos 3x = 1\/4<\/p>\n\n\n\n<p>4 cos x cos 2x cos 3x \u2013 1 = 0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>2 cos A cos B = cos (A + B) + cos (A \u2013 B)<\/p>\n\n\n\n<p>2(2cos x cos 3x) cos 2x \u2013 1 = 0<\/p>\n\n\n\n<p>2(cos 4x + cos 2x) cos2x \u2013 1 = 0<\/p>\n\n\n\n<p>2(2cos<sup>2<\/sup>&nbsp;2x \u2013 1 + cos 2x) cos 2x \u2013 1 = 0 [using cos 2A = 2cos<sup>2<\/sup>A \u2013 1]<\/p>\n\n\n\n<p>4cos<sup>3<\/sup>&nbsp;2x \u2013 2cos 2x + 2cos<sup>2<\/sup>&nbsp;2x \u2013 1 = 0<\/p>\n\n\n\n<p>2cos<sup>2<\/sup>&nbsp;2x (2cos 2x + 1) -1(2cos 2x + 1) = 0<\/p>\n\n\n\n<p>(2cos<sup>2<\/sup>&nbsp;2x \u2013 1) (2 cos 2x + 1) = 0<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>2cos 2x + 1 = 0 or (2cos<sup>2<\/sup>&nbsp;2x \u2013 1) = 0<\/p>\n\n\n\n<p>cos 2x = -1\/2 or cos 4x = 0 [using cos 2\u03b8 = 2cos<sup>2<\/sup>\u03b8 \u2013 1]<\/p>\n\n\n\n<p>cos 2x = cos (\u03c0 \u2013 \u03c0\/3) or cos 4x = cos \u03c0\/2<\/p>\n\n\n\n<p>cos 2x = cos 2\u03c0\/3 or cos 4x = cos \u03c0\/2<\/p>\n\n\n\n<p>2x = 2m\u03c0 \u00b1 2\u03c0\/3 or 4x = (2n + 1) \u03c0\/2<\/p>\n\n\n\n<p>x = m\u03c0 \u00b1 \u03c0\/3 or x = (2n + 1) \u03c0\/8<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = m\u03c0 \u00b1 \u03c0\/3 or (2n + 1) \u03c0\/8, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(v)&nbsp;<\/strong>cos x + sin x = cos 2x + sin 2x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>cos x + sin x = cos 2x + sin 2x<\/p>\n\n\n\n<p>upon rearranging we get,<\/p>\n\n\n\n<p>cos x \u2013 cos 2x = sin 2x \u2013 sin x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>cos A \u2013 cos B = \u2013 2 sin (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>-2 sin (2x+x)\/2 sin (2x-x)\/2 = 2 cos (2x+x)\/2 sin (2x-x)\/2<\/p>\n\n\n\n<p>2 sin 3x\/2 sin x\/2 = 2 cos 3x\/2 sin x\/2<\/p>\n\n\n\n<p>Sin x\/2 (sin 3x\/2 \u2013 cos 3x\/2) = 0<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>Sin x\/2 = 0 or sin 3x\/2 = cos 3x\/2<\/p>\n\n\n\n<p>Sin x\/2 = sin m\u03c0 or sin 3x\/2 \/ cos 3x\/2 = 0<\/p>\n\n\n\n<p>Sin x\/2 = sin m\u03c0 or tan 3x\/2 = 1<\/p>\n\n\n\n<p>Sin x\/2 = sin m\u03c0 or tan 3x\/2 = tan \u03c0\/4<\/p>\n\n\n\n<p>x\/2 = m\u03c0 or 3x\/2 = n\u03c0 + \u03c0\/4<\/p>\n\n\n\n<p>x = 2m\u03c0 or x = 2n\u03c0\/3 + \u03c0\/6<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = 2m\u03c0 or 2n\u03c0\/3 + \u03c0\/6, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(vi)&nbsp;<\/strong>sin x + sin 2x + sin 3x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin x + sin 2x + sin 3x = 0<\/p>\n\n\n\n<p>we shall rearrange and use transformation formula<\/p>\n\n\n\n<p>sin 2x + sin x + sin 3x = 0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>Sin 2x + 2 sin (3x+x)\/2 cos (3x-x)\/2 = 0<\/p>\n\n\n\n<p>Sin 2x + 2sin 2x cos x = 0<\/p>\n\n\n\n<p>Sin 2x (2 cos x + 1) = 0<\/p>\n\n\n\n<p>Sin 2x = 0 or 2cos x + 1 = 0<\/p>\n\n\n\n<p>Sin 2x = sin 0 or cos x = -1\/2<\/p>\n\n\n\n<p>Sin 2x = sin 0 or cos x = cos (\u03c0 \u2013 \u03c0\/3)<\/p>\n\n\n\n<p>Sin 2x = sin 0 or cos x = cos 2\u03c0\/3<\/p>\n\n\n\n<p>2x = n\u03c0 or x = 2m\u03c0 \u00b1 2\u03c0\/3<\/p>\n\n\n\n<p>x = n\u03c0\/2 or x = 2m\u03c0 \u00b1 2\u03c0\/3<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0\/2 or 2m\u03c0 \u00b1 2\u03c0\/3, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(vii)&nbsp;<\/strong>sin x + sin 2x + sin 3x + sin 4x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin x + sin 2x + sin 3x + sin 4x = 0<\/p>\n\n\n\n<p>we shall rearrange and use transformation formula<\/p>\n\n\n\n<p>sin x + sin 3x + sin 2x + sin 4x = 0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>2 sin (3x+x)\/2 cos (3x-x)\/2 + 2 sin (4x+2x)\/2 cos (4x-2x)\/2 = 0<\/p>\n\n\n\n<p>2 sin 2x cos x + 2 sin 3x cos x = 0<\/p>\n\n\n\n<p>2cos x (sin 2x + sin 3x) = 0<\/p>\n\n\n\n<p>Again by using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>we get,<\/p>\n\n\n\n<p>2cos x (2 sin (3x+2x)\/2 cos (3x-2x)\/2) = 0<\/p>\n\n\n\n<p>2cos x (2 sin 5x\/2 cos x\/2) = 0<\/p>\n\n\n\n<p>4 cos x sin 5x\/2 cos x\/2 = 0<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>Cos x = 0 or sin 5x\/2 = 0 or cos x\/2 = 0<\/p>\n\n\n\n<p>Cos x = cos 0 or sin 5x\/2 = sin 0 or cos x\/2 = cos 0<\/p>\n\n\n\n<p>Cos x = cos \u03c0\/2 or sin 5x\/2 = k\u03c0 or cos x\/2 = cos (2p + 1) \u03c0\/2<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/2 or 5x\/2 = k\u03c0 or x\/2 = (2p + 1) \u03c0\/2<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/2 or x = 2k\u03c0\/5 or x = (2p + 1)<\/p>\n\n\n\n<p>x = n\u03c0 + \u03c0\/2 or x = 2k\u03c0\/5 or x = (2p + 1)<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 + \u03c0\/2 or x = 2k\u03c0\/5 or x = (2p + 1), where n, k, p&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(viii)&nbsp;<\/strong>sin 3x \u2013 sin x = 4 cos<sup>2<\/sup>&nbsp;x \u2013 2<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin 3x \u2013 sin x = 4 cos<sup>2<\/sup>&nbsp;x \u2013 2<\/p>\n\n\n\n<p>sin 3x \u2013 sin x = 2(2 cos<sup>2<\/sup>&nbsp;x \u2013 1)<\/p>\n\n\n\n<p>sin 3x \u2013 sin x = 2 cos 2x [since,&nbsp;cos 2A = 2cos<sup>2<\/sup>&nbsp;A \u2013 1]<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>Sin A \u2013 sin B = 2 cos (A+B)\/2 sin (A-B)\/2<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>2 cos (3x+x)\/2 sin (3x-x)\/2 = 2 cos 2x<\/p>\n\n\n\n<p>2 cos 2x sin x \u2013 2 cos 2x = 0<\/p>\n\n\n\n<p>2 cos 2x (sin x \u2013 1) = 0<\/p>\n\n\n\n<p>Then,<\/p>\n\n\n\n<p>2 cos 2x = 0 or sin x \u2013 1 = 0<\/p>\n\n\n\n<p>Cos 2x = 0 or sin x = 1<\/p>\n\n\n\n<p>Cos 2x = cos 0 or sin x = sin 1<\/p>\n\n\n\n<p>Cos 2x = cos 0 or sin x = sin \u03c0\/2<\/p>\n\n\n\n<p>2x = (2n + 1) \u03c0\/2 or x = m\u03c0 + (-1)<sup>&nbsp;m<\/sup>&nbsp;\u03c0\/2<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/4 or x = m\u03c0 + (-1)<sup>&nbsp;m<\/sup>&nbsp;\u03c0\/2<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = (2n + 1) \u03c0\/4 or m\u03c0 + (-1)<sup>&nbsp;m<\/sup>&nbsp;\u03c0\/2, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>(ix)&nbsp;<\/strong>sin 2x \u2013 sin 4x + sin 6x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>sin 2x \u2013 sin 4x + sin 6x = 0<\/p>\n\n\n\n<p>we shall rearrange and use transformation formula<\/p>\n\n\n\n<p>\u2013 sin 4x + sin 6x + sin 2x = 0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>sin A + sin B = 2 sin (A+B)\/2 cos (A-B)\/2<\/p>\n\n\n\n<p>we get,<\/p>\n\n\n\n<p>\u2013 sin 4x + 2 sin (6x+2x)\/2 cos (6x-2x)\/2 = 0<\/p>\n\n\n\n<p>\u2013 sin 4x + 2 sin 4x cos 2x = 0<\/p>\n\n\n\n<p>Sin 4x (2 cos 2x \u2013 1) = 0<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>Sin 4x = 0 or 2 cos 2x \u2013 1 = 0<\/p>\n\n\n\n<p>Sin 4x = sin 0 or cos 2x = 1\/2<\/p>\n\n\n\n<p>Sin 4x = sin 0 or cos 2x = \u03c0\/3<\/p>\n\n\n\n<p>4x = n\u03c0 or 2x = 2m\u03c0 \u00b1 \u03c0\/3<\/p>\n\n\n\n<p>x = n\u03c0\/4 or x = m\u03c0 \u00b1 \u03c0\/6<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0\/4 or m\u03c0 \u00b1 \u03c0\/6, where m, n&nbsp;\u03f5&nbsp;Z.<\/p>\n\n\n\n<p><strong>5. Solve the following equations:<\/strong><\/p>\n\n\n\n<p><strong>(i) tan x + tan 2x + tan 3x = 0<\/strong><\/p>\n\n\n\n<p><strong>(ii) tan x + tan 2x = tan 3x<\/strong><\/p>\n\n\n\n<p><strong>(iii) tan 3x + tan x = 2 tan 2x<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>The general solution of any trigonometric equation is given as:<\/p>\n\n\n\n<p>sin x = sin y, implies x = n\u03c0 + (\u2013 1)<sup>n&nbsp;<\/sup>y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>cos x = cos y, implies x = 2n\u03c0&nbsp;\u00b1&nbsp;y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p>tan x = tan y, implies x = n\u03c0&nbsp;+ y, where n&nbsp;\u2208&nbsp;Z.<\/p>\n\n\n\n<p><strong>(i)&nbsp;<\/strong>tan x + tan 2x + tan 3x = 0<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan x + tan 2x + tan 3x = 0<\/p>\n\n\n\n<p>tan x + tan 2x + tan (x + 2x) = 0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>tan (A+B) = [tan A + tan B] \/ [1 \u2013 tan A tan B]<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>tan x + tan 2x + [[tan x + tan 2x]\/[1- tan x tan 2x]] = 0<\/p>\n\n\n\n<p>(tan x + tan 2x) (1 + 1\/(1- tan x tan 2x)) = 0<\/p>\n\n\n\n<p>(tan x + tan 2x) ([2 \u2013 tan x tan 2x] \/ [1 \u2013 tan x tan 2x]) = 0<\/p>\n\n\n\n<p>Then,<\/p>\n\n\n\n<p>(tan x + tan 2x) = 0 or ([2 \u2013 tan x tan 2x] \/ [1 \u2013 tan x tan 2x]) = 0<\/p>\n\n\n\n<p>(tan x + tan 2x) = 0 or [2 \u2013 tan x tan 2x] = 0<\/p>\n\n\n\n<p>tan x = tan (-2x) or tan x tan 2x = 2<\/p>\n\n\n\n<p>x = n\u03c0 + (-2x) or tax x [2tan x\/(1 \u2013 tan<sup>2<\/sup>&nbsp;x)] = 2 [Using, tan 2x = 2 tan x \/ 1-tan<sup>2<\/sup>&nbsp;x]<\/p>\n\n\n\n<p>3x = n\u03c0 or 2 tan<sup>2<\/sup>&nbsp;x \/ (1-tan<sup>2<\/sup>&nbsp;x) = 2<\/p>\n\n\n\n<p>3x = n\u03c0 or 2 tan<sup>2<\/sup>&nbsp;x = 2(1 \u2013 tan<sup>2<\/sup>&nbsp;x)<\/p>\n\n\n\n<p>3x = n\u03c0 or 2 tan<sup>2<\/sup>&nbsp;x = 2 \u2013 2tan<sup>2<\/sup>&nbsp;x<\/p>\n\n\n\n<p>3x = n\u03c0 or 4 tan<sup>2<\/sup>&nbsp;x = 2<\/p>\n\n\n\n<p>x = n\u03c0\/3 or tan<sup>2<\/sup>&nbsp;x = 2\/4<\/p>\n\n\n\n<p>x = n\u03c0\/3 or tan<sup>2<\/sup>&nbsp;x = 1\/2<\/p>\n\n\n\n<p>x = n\u03c0\/3 or tan x = 1\/<strong>\u221a<\/strong>2<\/p>\n\n\n\n<p>x = n\u03c0\/3 or x = tan \u03b1 [let 1\/<strong>\u221a<\/strong>2 be \u2018\u03b1\u2019]<\/p>\n\n\n\n<p>x = n\u03c0\/3 or x = m\u03c0 + \u03b1<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0\/3 or m\u03c0 + \u03b1, where \u03b1 = tan<sup>-1<\/sup>1\/<strong>\u221a<\/strong>2, m, n \u2208&nbsp;Z.<\/p>\n\n\n\n<p><strong>(ii)&nbsp;<\/strong>tan x + tan 2x = tan 3x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan x + tan 2x = tan 3x<\/p>\n\n\n\n<p>tan x + tan 2x \u2013 tan 3x = 0<\/p>\n\n\n\n<p>tan x + tan 2x \u2013 tan (x + 2x) = 0<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>tan (A+B) = [tan A + tan B] \/ [1 \u2013 tan A tan B]<\/p>\n\n\n\n<p>So,<\/p>\n\n\n\n<p>tan x + tan 2x \u2013 [[tan x + tan 2x]\/[1- tan x tan 2x]] = 0<\/p>\n\n\n\n<p>(tan x + tan 2x) (1 \u2013 1\/(1- tan x tan 2x)) = 0<\/p>\n\n\n\n<p>(tan x + tan 2x) ([\u2013 tan x tan 2x] \/ [1 \u2013 tan x tan 2x]) = 0<\/p>\n\n\n\n<p>Then,<\/p>\n\n\n\n<p>(tan x + tan 2x) = 0 or ([\u2013 tan x tan 2x] \/ [1 \u2013 tan x tan 2x]) = 0<\/p>\n\n\n\n<p>(tan x + tan 2x) = 0 or [\u2013 tan x tan 2x] = 0<\/p>\n\n\n\n<p>tan x = tan (-2x) or -tan x tan 2x = 0<\/p>\n\n\n\n<p>tan x = tan (-2x) or 2tan<sup>2<\/sup>&nbsp;x \/ (1 \u2013 tan<sup>2<\/sup>&nbsp;x) = 0 [Using, tan 2x = 2 tan x \/ 1-tan<sup>2<\/sup>&nbsp;x]<\/p>\n\n\n\n<p>x = n\u03c0 + (-2x) or x = m\u03c0 + 0<\/p>\n\n\n\n<p>3x = n\u03c0 or x = m\u03c0<\/p>\n\n\n\n<p>x = n\u03c0\/3 or x = m\u03c0<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0\/3 or m\u03c0, where m, n \u2208&nbsp;Z.<\/p>\n\n\n\n<p><strong>(iii)&nbsp;<\/strong>tan 3x + tan x = 2 tan 2x<\/p>\n\n\n\n<p>Let us simplify,<\/p>\n\n\n\n<p>tan 3x + tan x = 2 tan 2x<\/p>\n\n\n\n<p>tan 3x + tan x = tan 2x + tan 2x<\/p>\n\n\n\n<p>upon rearranging we get,<\/p>\n\n\n\n<p>tan 3x \u2013 tan 2x = tan 2x \u2013 tan x<\/p>\n\n\n\n<p>By using the formula,<\/p>\n\n\n\n<p>tan (A-B) = [tan A \u2013 tan B] \/ [1 + tan A tan B]<\/p>\n\n\n\n<p>so,[(tan 3x \u2013 tan 2x) (1+tan 3x tan 2x)] \/ [1 + tan 3x tan 2x] = [(tan 2x-tan x) (1+tan x tan 2x)] \/ [1 + tan 2x tan x]<\/p>\n\n\n\n<p>tan (3x \u2013 2x) (1 + tan 3x tan 2x) = tan (2x \u2013 x) (1 + tan x tan 2x)<\/p>\n\n\n\n<p>tan x [1 + tan 3x tan 2x \u2013 1 \u2013 tan 2x tan x] = 0<\/p>\n\n\n\n<p>tan x tan 2x (tan 3x \u2013 tan x) = 0<\/p>\n\n\n\n<p>so,<\/p>\n\n\n\n<p>tan x = 0 or tan 2x = 0 or (tan 3x \u2013 tan x) = 0<\/p>\n\n\n\n<p>tan x = 0 or tan 2x = 0 or tan 3x = tan x<\/p>\n\n\n\n<p>x = n\u03c0 or 2x = m\u03c0 or 3x = k\u03c0 + x<\/p>\n\n\n\n<p>x = n\u03c0 or x = m\u03c0\/2 or 2x = k\u03c0<\/p>\n\n\n\n<p>x = n\u03c0 or x = m\u03c0\/2 or x = k\u03c0\/2<\/p>\n\n\n\n<p>\u2234 the general solution is<\/p>\n\n\n\n<p>x = n\u03c0 or m\u03c0\/2 or k\u03c0\/2, where, m, n, k \u2208&nbsp;Z.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-11-maths-chapter-11-download-pdf\">RD Sharma Solutions for Class 11 Maths Chapter 11:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RD-Sharma-Solutions-for-Class-11-Maths-Chapter-11\u2013Trigonometric-Equations.pdf\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise RD Sharma Solutions for Class 11&nbsp;Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-1-sets\/\">Chapter 1\u2013Sets<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-2-relations\/\">Chapter 2\u2013Relations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-3-functions\/\">Chapter 3\u2013Functions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-4-measurement-of-angles\/\">Chapter 4\u2013Measurement of Angles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-5-trigonometric-functions\/\">Chapter 5\u2013Trigonometric Functions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-6-graphs-of-trigonometric-functions\/\">Chapter 6\u2013Graphs of Trigonometric Functions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-7-values-of-trigonometric-functions-at-sum-or-difference-of-angles\/\">Chapter 7\u2013Values of Trigonometric Functions at Sum or Difference of Angles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\">Chapter 8\u2013Transformation Formulae<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-9-values-of-trigonometric-functions-at-multiples-and-submultiples-of-an-angle\/\">Chapter 9\u2013Values of Trigonometric Functions at Multiples and Submultiples of an Angle<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-10-sine-and-cosine-formulae-and-their-applications\/\">Chapter 10\u2013Sine and Cosine Formulae and their Applications<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\">Chapter 11\u2013Trigonometric Equations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-12-mathematical-induction\/\">Chapter 12\u2013Mathematical Induction<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-13-complex-numbers\/\">Chapter 13\u2013Complex Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-14-quadratic-equations\/\">Chapter 14\u2013Quadratic Equations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-15-linear-inequations\/\">Chapter 15\u2013Linear Inequations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-16-permutations\/\">Chapter 16\u2013Permutations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-17-combinations\/\">Chapter 17\u2013Combinations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem\/\">Chapter 18\u2013Binomial Theorem<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-19-arithmetic-progressions\/\">Chapter 19\u2013Arithmetic Progressions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-20-geometric-progressions\/\">Chapter 20\u2013Geometric Progressions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-21-some-special-series\/\">Chapter 21\u2013Some Special Series<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-22-brief-review-of-cartesian-system-of-rectangular-coordinates\/\">Chapter 22\u2013Brief review of Cartesian System of Rectangular Coordinates<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-23-the-straight-lines\/\">Chapter 23\u2013The Straight Lines<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-24-the-circle\/\">Chapter 24\u2013The Circle<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-25-parabola\/\">Chapter 25\u2013Parabola<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-26-ellipse\/\">Chapter 26\u2013Ellipse<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-27-hyperbola\/\">Chapter 27\u2013Hyperbola<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-28-introduction-to-three-dimensional-coordinate-geometry\/\">Chapter 28\u2013Introduction to Three Dimensional Coordinate Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-29-limits\/\">Chapter 29\u2013Limits<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-30-derivatives\/\">Chapter 30\u2013Derivatives<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-31-mathematical-reasoning\/\">Chapter 31\u2013Mathematical Reasoning<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-32-statistics\/\">Chapter 32\u2013Statistics<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-33-probability\/\">Chapter 33\u2013Probability<\/a><\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">About RD Sharma<\/h2>\n\n\n\n<p>RD Sharma i<em>sn&#8217;t the kind of author you&#8217;d bump into at lit fests. But his bestselling books have helped many&nbsp;<\/em>CBSE<em>&nbsp;students lose their dread of&nbsp;<\/em>maths<em>. Sunday Times profiles the tutor turned internet star<\/em><br>He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like &#8216;series solution of linear differential equations&#8217;. Meet Dr&nbsp;Ravi Dutt Sharma&nbsp;\u2014&nbsp;mathematics&nbsp;teacher and author of 25 reference books \u2014 whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it&#8217;s only recently that a spoof video turned the tutor into a YouTube star.<\/p>\n\n\n\n<p>R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. &#8220;I like to spend all my time thinking and writing about maths problems. I find it relaxing,&#8221; he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government&#8217;s Guru Nanak Dev Institute of Technology.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Read More<\/h2>\n\n\n\n<ul class=\"wp-block-yoast-seo-related-links\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-9-values-of-trigonometric-functions-at-multiples-and-submultiples-of-an-angle\/\">RD Sharma Solutions for Class 11 Maths Chapter 9\u2013Values of Trigonometric Functions at Multiples and Submultiples of an Angle<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\">NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-12-maths-chapter-4-inverse-trigonometric-functions\/\">RD Sharma Solutions for Class 12 Maths Chapter 4\u2013Inverse Trigonometric Functions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-7-values-of-trigonometric-functions-at-sum-or-difference-of-angles\/\">RD Sharma Solutions for Class 11 Maths Chapter 7\u2013Values of Trigonometric Functions at Sum or Difference of Angles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-8-transformation-formulae\/\">RD Sharma Solutions for Class 11 Maths Chapter 8\u2013Transformation Formulae<\/a><\/li><\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Class 11: Maths Chapter 11 solutions. Complete Class 11 Maths Chapter 11 Notes. RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations RD Sharma 11th Maths Chapter 11, Class 11 Maths Chapter 11 solutions 1. Find the general solutions of the following equations: (i) sin x = 1\/2 (ii) cos x = \u2013 \u221a3\/2 [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":543388,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,919],"tags":[1962],"boards":[],"class_list":["post-543385","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-11","tag-rd-sharma-solutions","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>RD Sharma Solutions for Class 11, maths Chapter 11 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations | Browse Class 11 Maths Chapters RD Sharma - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations\" \/>\n<meta property=\"og:description\" content=\"Class 11: Maths Chapter 11 solutions. Complete Class 11 Maths Chapter 11 Notes. RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2021-09-30T04:19:40+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-10-01T06:45:00+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/i1.wp.com\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/class11m11.png?fit=1200%2C675&ssl=1\" \/>\n\t<meta property=\"og:image:width\" content=\"1200\" \/>\n\t<meta property=\"og:image:height\" content=\"675\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"24 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations\",\"datePublished\":\"2021-09-30T04:19:40+00:00\",\"dateModified\":\"2021-10-01T06:45:00+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\"},\"wordCount\":4170,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/class11m11.png\",\"keywords\":[\"RD Sharma Solutions\"],\"articleSection\":[\"Book Solutions\",\"class 11\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\",\"name\":\"RD Sharma Solutions for Class 11, maths Chapter 11 - 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