Class 11: Trigonometric Equations – Exercise 11.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the general solutions of the following equations:

$\displaystyle \text{i) } \sin x = \frac{1}{2} \hspace{1.0cm} \text{ii) } \cos x = - \frac{\sqrt{3}}{2} \hspace{1.0cm} \text{iii) } \mathrm{cosec} x = - \sqrt{2}$

$\displaystyle \text{iv) } \sec x = \sqrt{2} \hspace{1.0cm} \text{v) } \tan x = - \frac{1}{\sqrt{3}} \hspace{1.0cm} \text{vi) } \sqrt{3} \sec x = 2$

Answer:

$\displaystyle \text{i) } \sin x = \frac{1}{2}$

$\displaystyle \Rightarrow \sin x = \sin \frac{\pi}{6}$

$\displaystyle \Rightarrow x = n\pi + (-1)^n \frac{\pi}{6} ; n \in Z$

$\displaystyle \text{ii) } \cos x = - \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow \cos x = \cos ( \pi + \frac{\pi}{6} )$

$\displaystyle \Rightarrow \cos x = \cos \frac{7\pi}{6}$

$\displaystyle \Rightarrow x = 2n\pi \pm \frac{7\pi}{6} ; n \in Z$

$\displaystyle \text{iii) } \mathrm{cosec} x = - \sqrt{2}$

$\displaystyle \Rightarrow \frac{1}{\sin x} = -\sqrt{2}$

$\displaystyle \Rightarrow \sin x = - \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \sin x = \sin ( \pi + \frac{\pi}{4} )$

$\displaystyle \Rightarrow \sin x = \sin \frac{5\pi}{4}$

$\displaystyle \Rightarrow x = n\pi + (-1)^n \frac{\pi}{4} ; n \in Z$

$\displaystyle \text{iv) } \sec x = \sqrt{2}$

$\displaystyle \Rightarrow \frac{1}{\cos x} = \sqrt{2}$

$\displaystyle \Rightarrow \cos x = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi \pm \frac{\pi}{4} ; n \in Z$

$\displaystyle \text{v) } \tan x = - \frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow \tan x = - \tan \frac{\pi}{6} = \tan {\frac{-\pi}{6}}$

$\displaystyle \Rightarrow x = n\pi - \frac{\pi}{6} ; n \in Z$

$\displaystyle \text{vi) } \sqrt{3} \sec x = 2$

$\displaystyle \Rightarrow \frac{\sqrt{3}}{\cos x} = 2$

$\displaystyle \Rightarrow \cos x = \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{6}$

$\displaystyle \Rightarrow x = 2n\pi \pm \frac{\pi}{6} ; n \in Z$

$\displaystyle \\$

Question 2: Find the general solutions of the following equations:

$\displaystyle \text{i) } \sin 2x = \frac{\sqrt{3}}{2} \hspace{1.0cm} \text{ii) } \cos 3x = \frac{1}{2} \hspace{1.0cm} \text{iii) } \sin 9x = \sin x$

$\displaystyle \text{iv) } \sin 2x = \cos 3x \hspace{1.0cm} \text{v) } \tan x + \cot 2x = 0 \hspace{1.0cm} \text{vi) } \tan 3x = \cot x$

$\displaystyle \text{vii) } \tan 2x \tan x = 1 \hspace{1.0cm} \text{viii) } \tan mx + \cot nx = 0 \hspace{1.0cm} \text{ix) } \tan px = \cot qx$

$\displaystyle \text{x) } \sin 2x + \cos x = 0 \hspace{1.0cm} \text{xi) } \sin x = \tan x \hspace{1.0cm} \text{xii) } \sin 3x + \cos 2x = 0$

Answer:

$\displaystyle \text{i) } \sin 2x = \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow \sin 2x = \frac{\pi}{3}$

$\displaystyle \Rightarrow 2x = n\pi + ( -1)^n \frac{\pi}{3} ; n \in Z$

$\displaystyle \Rightarrow x = \frac{n\pi}{2} + ( -1)^n \frac{\pi}{6} ; n \in Z$

$\displaystyle \text{ii) } \cos 3x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos 3x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow 3x = 2n\pi \pm \frac{\pi}{3} ; n \in Z$

$\displaystyle \Rightarrow x = \frac{2n\pi}{3} \pm \frac{\pi}{9} ; n \in Z$

$\displaystyle \text{iii) } \sin 9x = \sin x$

$\displaystyle \Rightarrow \sin 9x - \sin x = 0$

$\displaystyle \Rightarrow 2 \cos 5x \sin x = 0$

Therefore, either

$\displaystyle \cos 5x = 0$

$\displaystyle \Rightarrow \cos 5x = \cos \frac{\pi}{2}$

$\displaystyle \Rightarrow 5x = (2n+1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = (2n+1) \frac{\pi}{10} ; n \in Z$

$\displaystyle \sin 4x = 0$

$\displaystyle \Rightarrow \sin 4x = \sin 0$

$\displaystyle \Rightarrow 4x = n\pi +(-1)^n ( 0)$

$\displaystyle \Rightarrow x = \frac{n\pi}{4} ; n \in Z$

$\displaystyle \text{iv) } \sin 2x = \cos 3x$

$\displaystyle \Rightarrow \cos ( \frac{\pi}{2} - 2x) = \cos 3x$

$\displaystyle \Rightarrow 3x = 2n\pi \pm ( \frac{\pi}{2} - 2x) ; n \in Z$

Considering positive sign

$\displaystyle 3x = 2n\pi + ( \frac{\pi}{2} - 2x) ; n \in Z$

$\displaystyle \Rightarrow 5x = ( 4n + 1 ) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = ( 4n + 1 ) \frac{\pi}{10} ; n \in Z$

Considering negative sign

$\displaystyle 3x = 2n\pi - ( \frac{\pi}{2} - 2x) ; n \in Z$

$\displaystyle \Rightarrow x = ( 4n - 1 ) \frac{\pi}{2} ; n \in Z$

$\displaystyle \text{v) } \tan x + \cot 2x = 0$

$\displaystyle \Rightarrow \tan x = - \cot 2x$

$\displaystyle \Rightarrow \tan 2x = - \cot x = - \tan ( \frac{\pi}{2} - x) = \tan ( x - \frac{\pi}{2} )$

$\displaystyle \Rightarrow 2x = n\pi + ( x - \frac{\pi}{2} ) ; n \in Z$

$\displaystyle \Rightarrow x = n\pi - \frac{\pi}{2} ; n \in Z$

$\displaystyle \text{vi) } \tan 3x = \cot x$

$\displaystyle \Rightarrow \tan 3x = \tan ( \frac{\pi}{2} - x)$

$\displaystyle \Rightarrow 3x = n\pi + ( \frac{\pi}{2} - x); n \in Z$

$\displaystyle \Rightarrow 4x = ( 2n+1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = ( 2n+1) \frac{\pi}{8} ; n \in Z$

$\displaystyle \text{vii) } \tan 2x \tan x = 1$

$\displaystyle \Rightarrow \tan 2x = \frac{1}{\tan x} = \cot x = \tan ( \frac{\pi}{2} -x)$

$\displaystyle \Rightarrow 2x = n\pi + ( \frac{\pi}{2} -x)$

$\displaystyle \Rightarrow 3x = ( 2n+1) \frac{\pi}{2}$

$\displaystyle \Rightarrow x = ( 2n+1) \frac{\pi}{6} ; n \in Z$

$\displaystyle \text{viii) } \tan mx + \cot nx = 0$

$\displaystyle \Rightarrow \tan mx = - \cot nx = \tan ( \frac{\pi}{2} + nx)$

$\displaystyle \Rightarrow mx = k\pi + ( nx + \frac{\pi}{2} ) ; k \in Z$

$\displaystyle \Rightarrow ( m-n) x = (2k+1) \frac{\pi}{2} ; k \in Z$

$\displaystyle x = (\frac{2k+1}{m-n})( \frac{\pi}{2}) ; k \in Z$

$\displaystyle \text{ix) } \tan px = \cot qx$

$\displaystyle \Rightarrow \tan px = \tan ( \frac{\pi}{2} - qx)$

$\displaystyle \Rightarrow px = n\pi + ( \frac{\pi}{2} - qx) ; n \in Z$

$\displaystyle \Rightarrow ( p+q) x = ( 2n+1 ) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = (\frac{2n+1}{p+q})( \frac{\pi}{2}) \; n \in Z$

$\displaystyle \text{x) } \sin 2x + \cos x = 0$

$\displaystyle \Rightarrow \cos x = - \sin 2x$

$\displaystyle \Rightarrow \cos x = - \cos ( \frac{\pi}{2} - 2x)$

$\displaystyle \Rightarrow \cos x = - \cos ( \pi - ( \frac{\pi}{2} - 2x))$

$\displaystyle \Rightarrow \cos x = \cos ( \frac{\pi}{2} + 2x)$

$\displaystyle \Rightarrow x = 2n\pi \pm ( \frac{\pi}{2} + 2x)$

Considering positive sign

$\displaystyle x = 2n\pi + ( \frac{\pi}{2} + 2x) ; n \in Z$

$\displaystyle \Rightarrow x = -2n\pi - \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = - ( 4n+1) \frac{\pi}{2} ; n \in Z$

Considering negative sign

$\displaystyle x = 2n\pi - ( \frac{\pi}{2} + 2x) ; n \in Z$

$\displaystyle \Rightarrow 3x = ( 4n - 1 ) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = (4n-1) \frac{\pi}{6} ; n \in Z$

$\displaystyle \text{xi) } \sin x = \tan x$

$\displaystyle \Rightarrow \sin x - \tan x = 0$

$\displaystyle \Rightarrow \sin x ( 1 - \frac{1}{\cos x} ) = 0$

Therefore, either

$\displaystyle \sin x = 0$

$\displaystyle \Rightarrow x = n\pi + ( -1)^n (0) ; n \in Z$

$\displaystyle \Rightarrow x = n\pi ; n \in Z$

$\displaystyle ( 1 - \frac{1}{\cos x} ) = 0$

$\displaystyle \Rightarrow \cos x -1 = 0$

$\displaystyle \Rightarrow x = 2m\pi ; m \in Z$

$\displaystyle \text{xii) } \sin 3x + \cos 2x = 0$

$\displaystyle \Rightarrow \cos 2x = - \sin 3x$

$\displaystyle \Rightarrow \cos 2x = - \cos ( \frac{\pi}{2} - 3x)$

$\displaystyle \Rightarrow \cos 2x = \cos ( \frac{\pi}{2} + 3x)$

$\displaystyle \Rightarrow 2x = 2n \pi \pm ( \frac{\pi}{2} + 3x); n \in Z$

Considering positive sign

$\displaystyle 2x = 2n \pi + ( \frac{\pi}{2} + 3x); n \in Z$

$\displaystyle \Rightarrow - x = ( 4n+1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = -( 4n+1) \frac{\pi}{2} ; n \in Z$

Considering negative sign

$\displaystyle 2x = 2n \pi - ( \frac{\pi}{2} + 3x); n \in Z$

$\displaystyle \Rightarrow 5x = ( 4n-1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = ( 4n-1) \frac{\pi}{10} ; n \in Z$

$\displaystyle \\$

Question 3: Solve the following equations:

$\displaystyle \text{i) } \sin^2 x - \cos x = \frac{1}{4} \hspace{1.0cm} \text{ii) } 2 \cos^2 x - 5 \cos x + 2 = 0$

$\displaystyle \text{iii) } 2 \sin^2 x + \sqrt{3} \cos x + 1 = 0 \hspace{1.0cm} \text{iv) } 4 \sin^2 x - 8 \cos x + 1 = 0$

$\displaystyle \text{v) } \tan^2 x + ( 1 - \sqrt{3}) \tan x - \sqrt{3} = 0 \hspace{1.0cm} \text{vi) } 3 \cos^2 x - 2 \sqrt{3} \sin x \cos x - 3 \sin^2 x = 0$

$\displaystyle \text{vii) } \cos 4x = \cos 2x$

Answer:

$\displaystyle \text{i) } \sin^2 x - \cos x = \frac{1}{4}$

$\displaystyle \Rightarrow 1 - \cos^2 x - \cos x = \frac{1}{4}$

$\displaystyle \Rightarrow \cos^2 x + \cos x - \frac{3}{4} = 0$

$\displaystyle \Rightarrow 4 \cos^2 x + 4 \cos x - 3 = 0$

$\displaystyle \Rightarrow 4 \cos^2 x + 6 \cos x - 2 \cos x - 3 = 0$

$\displaystyle \Rightarrow 2 \cos x ( 2 \cos x + 3) - ( 2 \cos x + 3) = 0$

$\displaystyle \Rightarrow ( 2 \cos x + 3)( 2\cos x - 1) = 0$

Therefore either

$\displaystyle 2 \cos x + 3 = 0$

$\displaystyle \Rightarrow \cos x = - \frac{3}{2}$

This is not possible as $\displaystyle -1 \leq \cos x \leq 1$

$\displaystyle 2 \cos x -1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow x = 2n\pi \pm \frac{\pi}{3} ; n \in Z$

$\displaystyle \text{ii) } 2 \cos^2 x - 5 \cos x + 2 = 0$

$\displaystyle 2 \cos^2 x - 4 \cos x - \cos x+2 = 0$

$\displaystyle \Rightarrow 2 \cos x ( \cos x - 2) - ( \cos x - 2) = 0$

$\displaystyle \Rightarrow (\cos x - 2) ( 2 \cos x - 1) = 0$

Therefore either

$\displaystyle \cos x - 2 = 0$

$\displaystyle \Rightarrow \cos x = 2$

This is not possible as $\displaystyle -1 \leq \cos x \leq 1$

$\displaystyle 2 \cos x -1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow x = 2n \pi \pm \frac{\pi}{3} ; n \in Z$

$\displaystyle \text{iii) } 2 \sin^2 x + \sqrt{3} \cos x + 1 = 0$

$\displaystyle \Rightarrow 2 - 2 \cos^2 x + \sqrt{3} \cos x + 1 = 0$

$\displaystyle \Rightarrow 2 \cos^2 x - \sqrt{3} \cos x -3 = 0$

$\displaystyle \Rightarrow 2 \cos^2 x -2 \sqrt{3} \cos x + \sqrt{3} \cos x -3 = 0$

$\displaystyle \Rightarrow 2 \cos x ( \cos x - \sqrt{3}) + \sqrt{3} ( \cos x - \sqrt{3}) = 0$

$\displaystyle \Rightarrow ( \cos x - \sqrt{3}) ( 2\cos x + \sqrt{3}) = 0$

Therefore either,

$\displaystyle \cos x - \sqrt{3} = 0$

$\displaystyle \Rightarrow \cos x = \sqrt{3}$

$\displaystyle -1 \leq \cos x \leq 1$

$\displaystyle not \ possible$

$\displaystyle 2\cos x + \sqrt{3}$

$\displaystyle \Rightarrow \cos x = - \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow \cos x = \cos ( \pi - \frac{\pi}{6} )$

$\displaystyle \Rightarrow \cos x = \cos \frac{5\pi}{6}$

$\displaystyle \Rightarrow x = 2n\pi \pm \frac{5\pi}{6} ; n \in Z$

$\displaystyle \text{iv) } 4 \sin^2 x - 8 \cos x + 1 = 0$

$\displaystyle \Rightarrow 4 - 4 \cos^2 x - 8 \cos x + 1 = 0$

$\displaystyle \Rightarrow 4 \cos^2 x + 8 \cos x -5 = 0$

$\displaystyle \Rightarrow 4 \cos^2 x + 10 \cos x - 2 \cos x -5 = 0$

$\displaystyle \Rightarrow 2 \cos x ( 2 \cos x + 5) - ( 2 \cos x + 5) = 0$

$\displaystyle \Rightarrow ( 2 \cos x + 5) ( 2 \cos x - 1) =0$

Therefore, either

$\displaystyle 2 \cos x + 5 = 0$

$\displaystyle \Rightarrow \cos x = - 2.5$

$\displaystyle Not \ possible$

$\displaystyle 2 \cos x -1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow x = 2n \pi \pm \frac{\pi}{3} ; n \in Z$

$\displaystyle \text{v) } \tan^2 x + ( 1 - \sqrt{3}) \tan x - \sqrt{3} = 0$

$\displaystyle \Rightarrow \tan^2 x + \tan x - \sqrt{3} \tan x - \sqrt{3} = 0$

$\displaystyle \Rightarrow \tan x ( \tan x + 1) - \sqrt{3} ( \tan x + 1) = 0$

$\displaystyle \Rightarrow (\tan x + 1) ( \tan x - \sqrt{3}) = 0$

Therefore, either

$\displaystyle \tan x + 1 = 0$

$\displaystyle \Rightarrow \tan x = - 1$

$\displaystyle \Rightarrow \tan x = - \tan \frac{\pi}{4} = \tan ( - \frac{\pi}{4} )$

$\displaystyle \Rightarrow x = n\pi - \frac{\pi}{4} ; n \in Z$

$\displaystyle \tan x - \sqrt{3} = 0$

$\displaystyle \Rightarrow \tan x = \sqrt{3}$

$\displaystyle \Rightarrow \tan x = \tan \frac{\pi}{3}$

$\displaystyle \Rightarrow x = m\pi + \frac{\pi}{3} ; m \in Z$

$\displaystyle \text{vi) } 3 \cos^2 x - 2 \sqrt{3} \sin x \cos x - 3 \sin^2 x = 0$

$\displaystyle \Rightarrow 3 \cos^2 x - 3 \sqrt{3} \sin x \cos x + \sqrt{3} \sin x \cos x - 3 \sin^2 x = 0$

$\displaystyle \Rightarrow 3 \cos x ( \cos x - \sqrt{3} \sin x) + \sqrt{3} \sin x ( \cos x - \sqrt{3} \sin x) = 0$

$\displaystyle \Rightarrow ( \cos x - \sqrt{3} \sin x)( 3 \cos x +\sqrt{3} \sin x) = 0$

Therefore, either

$\displaystyle \cos x - \sqrt{3} \sin x = 0$

$\displaystyle \Rightarrow \tan x = \frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow \tan x = \tan \frac{\pi}{6}$

$\displaystyle \Rightarrow x = n\pi + \frac{\pi}{6} ; n \in Z$

$\displaystyle 3 \cos x +\sqrt{3} \sin x = 0$

$\displaystyle \Rightarrow \tan x = - \sqrt{3}$

$\displaystyle \Rightarrow \tan x = \tan ( - \frac{\pi}{3} )$

$\displaystyle \Rightarrow x = m\pi - \frac{\pi}{3} ; m \in Z$

$\displaystyle \text{vii) } \cos 4x = \cos 2x$

$\displaystyle \Rightarrow \cos 4x- \cos 2x = 0$

$\displaystyle \Rightarrow 2 \sin 3x \sin x = 0$

Therefore, either

$\displaystyle \sin 3x = 0$

$\displaystyle \Rightarrow 3x = n\pi ; n \in Z$

$\displaystyle \sin x = 0$

$\displaystyle \Rightarrow x = m\pi ; m \in Z$

$\displaystyle \\$

Question 4: Solve the following equations:

$\displaystyle \text{i) } \cos x + \cos 2x + \cos 3x = 0 \hspace{1.0cm} \text{ii) } \cos x + \cos 3x - \cos 2x = 0$

$\displaystyle \text{iii) } \sin x + \sin 5x = \sin 3x \hspace{1.0cm} \text{iv) } \cos x \cos 2x \cos 3x = \frac{1}{4}$

$\displaystyle \text{v) } \cos x + \sin x = \cos 2x + \sin 2x \hspace{1.0cm} \text{vi) } \sin x + \sin 2x + \sin 3x = 0$

$\displaystyle \text{vii) } \sin x + \sin 2x + \sin 3x + \sin 4x = 0 \hspace{1.0cm} \text{viii) } \sin 3x - \sin x = 4 \cos^2 x - 2$

$\displaystyle \text{ix) } \sin 2x - \sin 4x + \sin 6x=0$

Answer:

$\displaystyle \text{i) } \cos x + \cos 2x + \cos 3x = 0$

$\displaystyle \Rightarrow \cos 2x + 2 \cos 2x \cos x = 0$

$\displaystyle \Rightarrow \cos 2x ( 1 + 2 \cos x) = 0$

Therefore either

$\displaystyle \cos 2x = 0$

$\displaystyle \Rightarrow \cos 2x = \cos \frac{\pi}{2}$

$\displaystyle \Rightarrow 2x = (2n+1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = (2n+1) \frac{\pi}{4} ; n \in Z$

$\displaystyle 1 + 2 \cos x = 0$

$\displaystyle \Rightarrow \cos x = - \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos ( \pi + \frac{\pi}{3} )$

$\displaystyle \Rightarrow \cos x = \cos \frac{4\pi}{3}$

$\displaystyle \Rightarrow x = m \pi \pm \frac{4\pi}{3} ; m \in Z$

$\displaystyle \text{ii) } \cos x + \cos 3x - \cos 2x = 0$

$\displaystyle \Rightarrow 2 \cos 2x \cos x - \cos 2x = 0$

$\displaystyle \Rightarrow \cos 2x ( 2 \cos x - 1) = 0$

Therefore, either

$\displaystyle \cos 2x = 0$

$\displaystyle \Rightarrow \cos 2x = \cos \frac{\pi}{2}$

$\displaystyle \Rightarrow 2x = (2n+1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = (2n+1) \frac{\pi}{4} ; n \in Z$

$\displaystyle 2 \cos x - 1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow x = 2m\pi \pm \frac{\pi}{3}$

$\displaystyle \text{iii) } \sin x + \sin 5x = \sin 3x$

$\displaystyle \Rightarrow 2 \sin 3x \cos 2x = \sin 3x$

$\displaystyle \Rightarrow \sin 3x ( 2 \cos 2x - 1) = 0$

Therefore,either

$\displaystyle \sin 3x = 0$

$\displaystyle \Rightarrow \sin 3x = \sin 0$

$\displaystyle \Rightarrow 3x = n\pi ; n \in Z$

$\displaystyle \Rightarrow x = \frac{n\pi}{3} ; n \in Z$

$\displaystyle 2 \cos 2x - 1 = 0$

$\displaystyle \Rightarrow \cos 2x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos 2x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow 2x = 2m\pi \pm \frac{\pi}{3}$

$\displaystyle \Rightarrow x = m\pi \pm \frac{\pi}{6}$

$\displaystyle \text{iv) } \cos x \cos 2x \cos 3x = \frac{1}{4}$

$\displaystyle (2 \cos x \cos 3x) \cos 2x = \frac{1}{2}$

$\displaystyle (\cos 4x + \cos x) \cos 2x = \frac{1}{2}$

$\displaystyle 2 \cos^3 2x + 2 \cos^2 2x - 2 \cos x = \frac{1}{2}$

$\displaystyle 4 \cos^3 2x + 2 \cos^2 2x - 2 \cos 2x - 1 = 0$

$\displaystyle 2 \cos 2x ( 2 \cos^2 2x -1 ) + ( 2 \cos^2 2x - 1)= 0$

$\displaystyle ( 2 \cos^2 2x - 1) ( 2 \cos 2x + 1) = 0$

Therefore, either

$\displaystyle 2 \cos^2 2x - 1 = 0$

$\displaystyle \Rightarrow \cos 4x = 0$

$\displaystyle \Rightarrow \cos 4x = \cos \frac{\pi}{2}$

$\displaystyle \Rightarrow 4x = ( 2n+1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = ( 2n+1) \frac{\pi}{8} ; n \in Z$

$\displaystyle 2 \cos 2x + 1 = 0$

$\displaystyle \Rightarrow \cos 2x = - \frac{1}{2}$

$\displaystyle \Rightarrow \cos 2x = \cos \frac{2\pi}{3}$

$\displaystyle \Rightarrow 2x = 2m\pi \pm \frac{2\pi}{3} ; n \in Z$

$\displaystyle \Rightarrow x = m\pi \pm \frac{\pi}{3} ; n \in Z$

$\displaystyle \text{v) } \cos x + \sin x = \cos 2x + \sin 2x$

$\displaystyle \Rightarrow \cos x - \cos 2x = \sin 2x - \sin x$

$\displaystyle \Rightarrow -2 \sin \frac{3x}{2} \sin \frac{-x}{2} = 2 \cos \frac{3x}{2} \sin \frac{x}{2}$

$\displaystyle \Rightarrow \sin \frac{3x}{2} \sin \frac{x}{2} = \cos \frac{3x}{2} \sin \frac{x}{2}$

$\displaystyle \Rightarrow \sin \frac{x}{2} \Big[ \sin \frac{3x}{2} - \cos \frac{3x}{2} \Big] = 0$

Therefore, either

$\displaystyle \sin \frac{x}{2} = 0$

$\displaystyle \Rightarrow \frac{x}{2} = n\pi ; n \in Z$

$\displaystyle \Rightarrow x = 2n\pi ; n \in Z$

$\displaystyle \sin \frac{3x}{2} - \cos \frac{3x}{2} = 0$

$\displaystyle \Rightarrow \tan \frac{3x}{2} = 1$

$\displaystyle \Rightarrow \tan \frac{3x}{2} = \tan \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{3x}{2} = m\pi + \frac{\pi}{4}$

$\displaystyle \Rightarrow x= \frac{2m\pi}{3} + \frac{\pi}{6} ; m \in Z$

$\displaystyle \text{vi) } \sin x + \sin 2x + \sin 3x = 0$

$\displaystyle \Rightarrow \sin 2x + 2 \sin 2x \cos x = 0$

$\displaystyle \Rightarrow \sin 2x ( 1 + 2 \cos 2x) = 0$

Therefore, either

$\displaystyle \sin 2x = 0$

$\displaystyle \Rightarrow 2x = n\pi ; n \in Z$

$\displaystyle \Rightarrow x = \frac{n\pi}{2} ; n \in Z$

$\displaystyle 1 + 2 \cos x = 0$

$\displaystyle \Rightarrow \cos x = \frac{-1}{2} = \cos \frac{2\pi}{3}$

$\displaystyle \Rightarrow x = 2m\pi \pm \frac{2\pi}{3} ; m \in Z$

$\displaystyle \text{vii) } \sin x + \sin 2x + \sin 3x + \sin 4x = 0$

$\displaystyle \Rightarrow ( \sin 2x + \sin 4x) + ( \sin x + \sin 3x) = 0$

$\displaystyle \Rightarrow 2 \sin 3x \cos x + 2 \sin 2x \cos x = 0$

$\displaystyle \Rightarrow \cos x ( \sin 3x + \sin 2x) = 0$

$\displaystyle \Rightarrow \cos x ( 2 \sin \frac{5x}{2} \cos \frac{x}{2} ) = 0$

Therefore, either

$\displaystyle \cos x = 0$

$\displaystyle \Rightarrow x = ( 2n + 1 ) \frac{\pi}{2} ; n \in Z$ or

$\displaystyle \sin \frac{5x}{2} = 0$

$\displaystyle \Rightarrow \frac{5x}{2} = m\pi ; m \in Z$

$\displaystyle \Rightarrow x = \frac{2m\pi}{5} ; m \in Z$

$\displaystyle \cos \frac{x}{2} = 0$

$\displaystyle \Rightarrow \frac{x}{2} = ( 2p+1) \frac{\pi}{2} ; p \in Z$

$\displaystyle \text{viii) } \sin 3x - \sin x = 4 \cos^2 x - 2$

$\displaystyle \Rightarrow 2 \cos 2x \sin x = 2 ( 2 \cos^2 x - 1)$

$\displaystyle \Rightarrow 2 \cos 2x \sin x = 2 \cos 2x$

$\displaystyle \Rightarrow \cos 2x ( \sin x - 1 ) = 0$

Therefore either,

$\displaystyle \cos 2x = 0$

$\displaystyle \Rightarrow 2x = ( 2n+1) \frac{\pi}{2} ; n \in Z$

$\displaystyle \Rightarrow x = ( 2n+1) \frac{\pi}{4} ; n \in Z$

$\displaystyle \sin x - 1 = 0$

$\displaystyle \Rightarrow x = m\pi + ( - 1 )^m \frac{\pi}{2} ; m \in Z$

$\displaystyle \text{ix) } \sin 2x - \sin 4x + \sin 6x=0$

$\displaystyle \Rightarrow (\sin 2x + \sin 6x) - \sin 4x = 0$

$\displaystyle \Rightarrow 2 \sin 4x \cos 2x - \sin 4x = 0$

$\displaystyle \Rightarrow \sin 4x ( 2 \cos 2x - 1 ) = 0$

Therefore either,

$\displaystyle \sin 4x = 0$

$\displaystyle \Rightarrow 4x = n\pi$

$\displaystyle \Rightarrow x = \frac{n\pi}{4} ; n \in Z$

$\displaystyle 2 \cos 2x - 1 = 0$

$\displaystyle \Rightarrow \cos 2x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos 2x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow 2x = 2m\pi \pm \frac{\pi}{3}$

$\displaystyle \Rightarrow x = m\pi \pm \frac{\pi}{6}$

$\displaystyle \\$

Question 5: Solve the following equations:

$\displaystyle \text{i) } \tan x + \tan 2x + \tan 3x = 0 \hspace{1.0cm} \text{ii) } \tan x + \tan 2x = \tan 3x$

$\displaystyle \text{iii) } \tan 3x + \tan x = 2 \tan 2x$

Answer:

$\displaystyle \text{i) } \tan x + \tan 2x + \tan 3x = 0$

$\displaystyle \Rightarrow \tan x + \tan 2x + \frac{\tan x + \tan 2x }{1 - \tan x \tan 2x} = 0$

$\displaystyle \Rightarrow (\tan x + \tan 2x) \Big[ 1 + \frac{1}{1 - \tan x \tan 2x} \Big] = 0$

$\displaystyle \Rightarrow (\tan x + \tan 2x) ( 2 - \tan x \tan 2x) = 0$

Therefore either,

$\displaystyle \tan x + \tan 2x = 0$

$\displaystyle \Rightarrow \tan x = - \tan 2x$

$\displaystyle \Rightarrow \tan x = \tan ( -2x)$

$\displaystyle \Rightarrow x = n\pi + ( - 2x) ; n \in Z$

$\displaystyle \Rightarrow 3x = n\pi ; n \in Z$

$\displaystyle \Rightarrow x = \frac{n\pi}{3} ; n \in Z$

$\displaystyle 2 - \tan x \tan 2x = 0$

$\displaystyle \Rightarrow \tan x \frac{2\tan x }{1 - \tan^2 x} = 2$

$\displaystyle \Rightarrow 2 \tan^2 x = 2 - 2 \tan^2 x$

$\displaystyle \Rightarrow 4 \tan^2 x = 2$

$\displaystyle \Rightarrow \tan^2 x = \frac{1}{2}$

$\displaystyle \Rightarrow \tan x = \pm \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow x = m\pi \pm \tan^{-1} \frac{1}{\sqrt{2}}$

$\displaystyle \text{ii) } \tan x + \tan 2x = \tan 3x$

$\displaystyle \Rightarrow \tan x + \tan 2x - \frac{\tan x + \tan 2x }{1 - \tan x \tan 2x} = 0$

$\displaystyle \Rightarrow (\tan x + \tan 2x) \Big[ 1 - \frac{1 }{1 - \tan x \tan 2x} \Big] = 0$

$\displaystyle \Rightarrow (\tan x + \tan 2x) \Big[ \frac{-\tan x \tan 2x }{1 - \tan x \tan 2x} \Big] = 0$

$\displaystyle \Rightarrow \tan x + \tan 2x = 0$

$\displaystyle \Rightarrow \tan x + \frac{2 \tan x }{1 - \tan^2 x } = 0$

$\displaystyle \Rightarrow \tan x \Big[ 1 + \frac{2}{1-\tan^2 x} \Big] = 0$

$\displaystyle \Rightarrow \tan x \Big[ \frac{3 - \tan^2 x}{1-\tan^2 x} \Big] = 0$

Therefore either,

$\displaystyle \tan x = 0$

$\displaystyle \Rightarrow x = m\pi ; m \in Z$

or $\displaystyle \tan^2 x = 3$

$\displaystyle \Rightarrow \tan x = \pm \sqrt{3}$

$\displaystyle \Rightarrow \tan x = \pm \tan \frac{\pi}{3}$

$\displaystyle \Rightarrow x = n\pi \pm \frac{\pi}{3} ; n \in Z$

or $\displaystyle \tan 2x = 0$

$\displaystyle \Rightarrow 2x = r \pi ; r \in Z$

$\displaystyle \Rightarrow x = \frac{r\pi}{2} ; r \in Z$

$\displaystyle \text{iii) } \tan 3x + \tan x = 2 \tan 2x$

$\displaystyle \Rightarrow \tan 3x - \tan 2x = \tan 2x - \tan x$

$\displaystyle \Rightarrow \tan x ( 1 + \tan 3x \tan 2x) = \tan x ( 1 + \tan 2x \tan x)$

$\displaystyle \Rightarrow \tan x + \tan x \tan 2x \tan 3x = \tan x + \tan 2x \tan^2 x$

$\displaystyle \Rightarrow \tan x \tan 2x ( \tan 3x - \tan x) = 0$

Therefore either,

$\displaystyle \tan x = 0$

$\displaystyle \Rightarrow x = n\pi ; n \in Z$

$\displaystyle \tan 2x = 0$

$\displaystyle \Rightarrow 2x = m\pi ; m \in Z$

$\displaystyle \Rightarrow x = \frac{m\pi}{2} ; m \in Z$ or

$\displaystyle \tan 3x - \tan x = 0$

$\displaystyle \Rightarrow \tan 3x = \tan x$

$\displaystyle \Rightarrow 3x = r\pi + x ; r \in Z$

$\displaystyle \Rightarrow 2x = r \pi ; r \in Z$

$\displaystyle \Rightarrow x = \frac{r\pi}{2} ; r \in Z$

$\displaystyle \\$

Question 6: Solve the following equations:

$\displaystyle \text{i) } \sin x + \cos x = \sqrt{2} \hspace{1.0cm} \text{ii) } \sqrt{3} \cos x + \sin x = 1 \hspace{1.0cm} \text{iii) } \sin x + \cos x = 1$

$\displaystyle \text{iv) } \mathrm{cosec} x = 1 + \cot x \hspace{1.0cm} \text{v) } (\sqrt{3} - 1) \cos x + (\sqrt{3}+1) \sin x = 2$

Answer:

$\displaystyle \text{i) } \sin x + \cos x = \sqrt{2}$

$\displaystyle \Rightarrow \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = 1$

$\displaystyle \Rightarrow \sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4} \cos x = 1$

$\displaystyle \Rightarrow \cos ( x - \frac{\pi}{4} ) = \cos 0$

$\displaystyle \Rightarrow x - \frac{\pi}{4} = 2n\pi$

$\displaystyle \Rightarrow x = 2n\pi + \frac{\pi}{4} ; n \in Z$

$\displaystyle \text{ii) } \sqrt{3} \cos x + \sin x = 1$

$\displaystyle \Rightarrow \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos \frac{\pi}{6} \cos x + \sin \frac{\pi}{6} \sin x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos ( x - \frac{\pi}{6} ) = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow x - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{3}$

Therefore either,

For positive sign

$\displaystyle x - \frac{\pi}{6} = 2n\pi + \frac{\pi}{3}$

$\displaystyle \Rightarrow x = 2n\pi + \frac{\pi}{2} ; n \in Z$

For negative sign

$\displaystyle x - \frac{\pi}{6} = 2n\pi - \frac{\pi}{3}$

$\displaystyle \Rightarrow x = 2n\pi - \frac{\pi}{6} ; n \in Z$

$\displaystyle \text{iii) } \sin x + \cos x = 1$

$\displaystyle \Rightarrow \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \cos ( x - \frac{\pi}{4} ) = \cos \frac{\pi}{4}$

$\displaystyle \Rightarrow x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}$

Therefore either,

For positive sign

$\displaystyle x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi + \frac{\pi}{2} ; n \in Z$

For negative sign

$\displaystyle x - \frac{\pi}{4} = 2n\pi - \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi ; n \in Z$

$\displaystyle \text{iv) } \mathrm{cosec} x = 1 + \cot x$

$\displaystyle \Rightarrow \frac{1}{\sin x} = 1 + \frac{\cos x}{\sin x}$

$\displaystyle \Rightarrow \sin x + \cos x = 1$

$\displaystyle \Rightarrow \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \cos ( x - \frac{\pi}{4} ) = \cos \frac{\pi}{4}$

$\displaystyle \Rightarrow x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}$

Therefore either,

For positive sign

$\displaystyle x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi + \frac{\pi}{2} ; n \in Z$

For negative sign

$\displaystyle x - \frac{\pi}{4} = 2n\pi - \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi ; n \in Z$

$\displaystyle \text{v) } (\sqrt{3} - 1) \cos x + (\sqrt{3}+1) \sin x = 2$

Divide by $\displaystyle \sqrt{( \sqrt{3}-1)^2 + ( \sqrt{3}+1)^2} = 2\sqrt{2}$

$\displaystyle \Rightarrow \frac{\sqrt{3}-1}{2} \cos x + \frac{\sqrt{3}+1}{2} \sin x = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \Big( \cos \frac{\pi}{6} \cos \frac{\pi}{4} - \sin \frac{\pi}{6} \sin \frac{\pi}{4} \Big) \cos x + \Big( \sin \frac{\pi}{6} \cos \frac{\pi}{4} + \cos \frac{\pi}{6} \sin \frac{\pi}{4} \Big) \sin x = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \cos x \cos ( \frac{\pi}{6} + \frac{\pi}{4} ) + \sin x \sin ( \frac{\pi}{6} + \frac{\pi}{4} ) = \cos \frac{\pi}{4}$

$\displaystyle \Rightarrow \cos x \cos \frac{5\pi}{12} + \sin x \sin \frac{5\pi}{12} = \cos \frac{\pi}{4}$

$\displaystyle \Rightarrow \cos ( x - \frac{5\pi}{12} ) = \cos \frac{\pi}{4}$

Therefore either,

For positive sign

$\displaystyle x - \frac{5\pi}{12} = 2n\pi + \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi + \frac{2\pi}{3} ; n \in Z$

For negative sign

$\displaystyle x - \frac{5\pi}{12} = 2n\pi - \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi + \frac{\pi}{6} ; n \in Z$

$\displaystyle \\$

Question 7: Solve the following equations:

$\displaystyle \text{i) } \cot x + \tan x = 2 \hspace{1.0cm} \text{ii) } 2 \sin^2 x = 3 \cos x, 0 \leq x \leq 2\pi$

$\displaystyle \text{iii) } \sec x \cos 5x + 1 = 0, 0 < x < \frac{\pi}{2} \hspace{1.0cm} \text{iv) } 5 \cos^2 x + 7 \sin^2 x - 6 =0$

$\displaystyle \text{v) } \sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$

$\displaystyle \text{vi) } 4 \sin x \cos x + 2 \sin x + 2 \cos x + 1 = 0 \hspace{1.0cm} \text{vii) } \cos x + \sin x = \cos 2x + \sin 2x$

$\displaystyle \text{viii) } \sin x \tan x - 1 = \tan x - \sin x$ ix) $\displaystyle 3 \tan x + \cot x = 5 \mathrm{cosec} x$

Answer:

$\displaystyle \text{i) } \cot x + \tan x = 2$

$\displaystyle \Rightarrow \frac{\cos x}{\sin x } + \frac{\sin x }{\cos x } = 2$

$\displaystyle \Rightarrow \cos^2 x + \sin^2 x = 2 \sin x \cos x$

$\displaystyle \Rightarrow \sin 2x = 1$

$\displaystyle \Rightarrow 2x = ( \frac{2n+1}{2} ) \pi ; n \in Z$

$\displaystyle \Rightarrow x = ( \frac{2n+1}{4} ) \pi ; n \in Z$

$\displaystyle \text{ii) } 2 \sin^2 x = 3 \cos x, 0 \leq x \leq 2\pi$

$\displaystyle \Rightarrow 2 - 2 \cos^2 x = 3 \cos x$

$\displaystyle \Rightarrow 2 \cos^2 x + 3 \cos x - 2 = 0$

$\displaystyle \Rightarrow ( \cos x + 2) ( 2 \cos x - 1) = 0$

Therefore either,

$\displaystyle \cos x + 2 = 0$

$\displaystyle \Rightarrow \cos x = - 2$

$\displaystyle Not \ possible$

$\displaystyle 2 \cos x -1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow x = \frac{\pi}{3} , \frac{5\pi}{3}$

$\displaystyle \text{iii) } \sec x \cos 5x + 1 = 0, 0 < x < \frac{\pi}{2}$

$\displaystyle \Rightarrow \frac{\cos 5x + \cos x }{\cos x } = 0$

This means that $\displaystyle \cos x \neq 0 \Rightarrow 2 \cos 3x \cos 2x = 0$

Therefore either,

$\displaystyle \Rightarrow \cos 3x = 0$

$\displaystyle \Rightarrow 3x = \frac{\pi}{2}$

$\displaystyle \Rightarrow x = \frac{\pi}{6}$

$\displaystyle \Rightarrow \cos 2x = 0$

$\displaystyle \Rightarrow 2x = \frac{\pi}{2}$

$\displaystyle \Rightarrow x = \frac{\pi}{4}$

$\displaystyle \text{iv) } 5 \cos^2 x + 7 \sin^2 x - 6 =0$

$\displaystyle \Rightarrow 5 - 5 \sin^2 x + 7 \sin^2 x - 6 = 0$

$\displaystyle \Rightarrow 2 \sin^2 x - 1 = 0$

$\displaystyle \Rightarrow \sin^2 x = \frac{1}{2}$

$\displaystyle \Rightarrow \sin x = \pm \frac{1}{2}$

$\displaystyle \Rightarrow \sin x = \pm \sin ( \frac{\pi}{4})$

$\displaystyle \Rightarrow x = n\pi \pm \frac{\pi}{4}$

$\displaystyle \text{v) } \sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$

$\displaystyle \Rightarrow (\sin x + \sin 3x) - 3 \sin 2x= ( \cos x + \cos 3x) - 3 \cos 2x$

$\displaystyle \Rightarrow 2 \sin 2x \cos x - 3 \sin 2x = 2 \cos 2x \cos x - 3 \cos 2x$

$\displaystyle \Rightarrow \sin 2x ( 2 \cos x - 3) = \cos 2x ( 2 \cos x - 3)$

$\displaystyle \Rightarrow (2 \cos x - 3) ( \sin 2x - \cos 2x) = 0$

Therefore either,

$\displaystyle 2\cos x - 3 = 0$

$\displaystyle \Rightarrow \cos x = \frac{3}{2}$

$\displaystyle Not \ possible$

$\displaystyle \sin 2x - \cos 2x = 0$

$\displaystyle \Rightarrow \tan 2x = 1$

$\displaystyle \Rightarrow \tan 2x = \tan \frac{\pi}{4}$

$\displaystyle \Rightarrow 2x = n\pi + \frac{\pi}{4}$

$\displaystyle \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{8}$

$\displaystyle \text{vi) } 4 \sin x \cos x + 2 \sin x + 2 \cos x + 1 = 0$

$\displaystyle \Rightarrow 2 \sin x ( 2 \cos x + 1 ) + ( 2 \cos x + 1) = 0$

$\displaystyle \Rightarrow ( 2 \cos x + 1)(2 \sin x + 1) = 0$

Therefore either,

$\displaystyle 2 \cos x + 1 = 0$

$\displaystyle \Rightarrow \cos x = - \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{2\pi}{3}$

$\displaystyle \Rightarrow x = 2n\pi \pm \frac{2\pi}{3} ; n \in Z$

$\displaystyle 2 \sin x + 1 = 0$

$\displaystyle \Rightarrow \sin x = - \frac{1}{2}$

$\displaystyle \Rightarrow \sin x = \sin ( \pi + \frac{\pi}{6})$

$\displaystyle \Rightarrow x = m\pi + ( -1)^m \frac{7\pi}{6} ; m \in Z$

$\displaystyle \text{vii) } \cos x + \sin x = \cos 2x + \sin 2x$

$\displaystyle \Rightarrow \cos x - \cos 2x = \sin 2x - \sin x$

$\displaystyle \Rightarrow 2 \sin \frac{3x}{2} \sin \frac{x}{2} = 2 \cos \frac{3x}{2} \sin \frac{x}{2}$

$\displaystyle \Rightarrow \sin \frac{x}{2} \Big[sin \frac{3x}{2} - \cos \frac{3x}{2} \Big] = 0$

Therefore either,

$\displaystyle \Rightarrow \sin \frac{x}{2} = 0$

$\displaystyle \Rightarrow \frac{x}{2} = n\pi ; n \in Z$

$\displaystyle \Rightarrow x = 2 n\pi ; n \in Z$

$\displaystyle \Rightarrow \tan \frac{3x}{2} = 1$

$\displaystyle \Rightarrow \tan \frac{3x}{2} = \tan \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{3x}{2} = m\pi + \frac{\pi}{4} ; m \in Z$

$\displaystyle \Rightarrow x = \frac{2m\pi}{3} + \frac{\pi}{6} ; m \in Z$

$\displaystyle \text{viii) } \sin x \tan x - 1 = \tan x - \sin x$

$\displaystyle \Rightarrow \sin x \tan x - \tan x - 1 + \sin x = 0$

$\displaystyle \Rightarrow \tan x ( \sin x - 1 ) + ( \sin x - 1) = 0$

$\displaystyle \Rightarrow ( \sin x - 1 ) ( \tan x + 1) = 0$

Therefore either,

$\displaystyle \sin x - 1 = 0$

$\displaystyle \Rightarrow \sin x = \sin \frac{\pi}{2}$

$\displaystyle \Rightarrow x = n\pi + ( -1 )^n \frac{\pi}{2}$

$\displaystyle \tan x + 1 = 0$

$\displaystyle \Rightarrow \tan x = - 1$

$\displaystyle \Rightarrow \tan x = \tan ( \frac{3\pi}{4} )$

$\displaystyle \Rightarrow x = m\pi + \frac{3\pi}{4}$

$\displaystyle \text{ix) } 3 \tan x + \cot x = 5 \mathrm{cosec} x$

$\displaystyle \Rightarrow 3 \frac{\sin x }{\cos x} + \frac{\cos x }{\sin x} = \frac{5}{\sin x}$

$\displaystyle \Rightarrow 3 - 3 \cos^2 x + \cos^2 x = 5 \cos x$

$\displaystyle \Rightarrow 2 \cos^2 x + 5 \cos x - 3 = 0$

$\displaystyle \Rightarrow 2 \cos^2 x + 6 \cos x - \cos x - 3 = 0$

$\displaystyle \Rightarrow 2 \cos x ( \cos x + 3) - ( \cos x + 3) = 0$

Therefore either,

$\displaystyle \Rightarrow \cos x + 3 = 0$

$\displaystyle \Rightarrow \cos x = - 3$

$\displaystyle Not \ possible$

$\displaystyle \Rightarrow 2 \cos x - 1 = 0$

$\displaystyle \Rightarrow \cos x = \frac{1}{2}$

$\displaystyle \Rightarrow \cos x = \cos \frac{\pi}{3}$

$\displaystyle \Rightarrow x = 2n\pi \pm \frac{\pi}{3} ; n \in Z$

$\displaystyle \\$

Solve the following equations:

Question 8: $\displaystyle 3 - 2 \cos x - 4 \sin x - \cos 2x + \sin 2x = 0$

Answer:

$\displaystyle 3 - 2 \cos x - 4 \sin x - \cos 2x + \sin 2x = 0$

$\displaystyle \Rightarrow 3 - \cos 2x - 4 \sin x + \sin 2x - 2 \cos x = 0$

$\displaystyle \Rightarrow 3 - ( 1- 2 \sin^2 x ) - 4 \sin x + 2 \sin x \cos x - 2 \cos x = 0$

$\displaystyle \Rightarrow 2 \sin^2 x + 2 - 4 \sin x + 2 \cos x ( \sin x - 1 ) = 0$

$\displaystyle \Rightarrow 2 ( \sin x - 1 )^2 + 2 \cos x ( \sin x - 1 ) = 0$

$\displaystyle \Rightarrow ( \sin x - 1 ) [ ( \sin x - 1 ) + \cos x ] = 0$

Therefore either,

$\displaystyle \sin x - 1 = 0$

$\displaystyle \Rightarrow \sin x = 1$

$\displaystyle \Rightarrow \sin x = \sin \frac{\pi}{2}$

$\displaystyle \Rightarrow x = m\pi + ( -1 )^n \frac{\pi}{2} ; m \in Z$

$\displaystyle \sin x - 1 + \cos x = 0$

$\displaystyle \Rightarrow \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \cos x \cos \frac{\pi}{4} + \sin x \sin \frac{\pi}{4} = \cos \frac{\pi}{4}$

$\displaystyle \Rightarrow \cos ( x - \frac{\pi}{4} ) = \cos \frac{\pi}{4}$

$\displaystyle \Rightarrow x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}$

With positive sign

$\displaystyle x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi + \frac{\pi}{2} ; n \in Z$

With negative sign

$\displaystyle x - \frac{\pi}{4} = 2n\pi - \frac{\pi}{4}$

$\displaystyle \Rightarrow x = 2n\pi ; n \in Z$

$\displaystyle \\$

Question 9: $\displaystyle 3 \sin^2 x - 5 \sin x \cos x + 8 \cos^2 x = 2$

Answer:

$\displaystyle 3 \sin^2 x - 5 \sin x \cos x + 8 \cos^2 x = 2$

$\displaystyle \Rightarrow 3 ( \sin^2 x + \cos^2 x ) - 5 \sin x \cos x + 5 \cos^2 x - 2 = 0$

$\displaystyle \Rightarrow 3 - 5 \sin x \cos x + 5 \cos^2 x - 2 = 0$

$\displaystyle \Rightarrow 5 \cos^2 x - 5 \sin x \cos x + 1 = 0$

$\displaystyle \Rightarrow 5 - 5 \sin^2 x - 5 \sin x \cos x + 1 = 0$

$\displaystyle \Rightarrow 5 \sin^2 x + 5 \sin x \cos x - 6 = 0$

Dividing by $\displaystyle \Rightarrow \cos^2 x$

$\displaystyle \Rightarrow 5 \tan^2 x + 5 \tan x - 6 \sec^2 x = 0$

$\displaystyle \Rightarrow 5 \tan^2 x + 5 \tan x - 6 (1 + \tan^2 x) = 0$

$\displaystyle \Rightarrow \tan^2 x - 5 \tan x + 6 = 0$

$\displaystyle \Rightarrow \tan^2 x - 3 \tan x - 2 \tan x + 6 = 0$

$\displaystyle \Rightarrow \tan x ( \tan x - 3) - 2 ( \tan x - 3) = 0$

Therefore either,

$\displaystyle \Rightarrow \tan x - 3 = 0$

$\displaystyle \Rightarrow \tan x = 3$

$\displaystyle \Rightarrow x = n\pi + \tan^{-1} 3; n \in Z$

$\displaystyle \Rightarrow \tan x - 2$

$\displaystyle \Rightarrow \tan x = 2$

$\displaystyle \Rightarrow x = m\pi + \tan^{-1} 2 ; m \in Z$

$\displaystyle \\$

Question 10: $\displaystyle 2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}$

Answer:

$\displaystyle 2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}$

$\displaystyle \Rightarrow 2^{\sin^2 x} + 2^{1 - \sin^2 x} = 2\sqrt{2}$

$\displaystyle \text{If } \Rightarrow 2^{\sin^2 x} = y$

$\displaystyle \Rightarrow y + \frac{2}{y} = 2\sqrt{2}$

$\displaystyle \Rightarrow y^2 - 2\sqrt{2} y + 2 = 0$

$\displaystyle \Rightarrow ( y - \sqrt{2})^2 = 0$

$\displaystyle \Rightarrow y = \sqrt{2}$

$\displaystyle \Rightarrow 2^{\sin^2 x} = 2^{\frac{1}{2}}$

$\displaystyle \Rightarrow \sin^2 x = \frac{1}{2}$

$\displaystyle \Rightarrow \sin x = \pm \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \sin x = \pm \sin \frac{\pi}{4}$

$\displaystyle \Rightarrow x = n\pi \pm \frac{\pi}{4} ; n \in Z$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 11: Trigonometric Equations – Exercise 11.1

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