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

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

iv) $\sec x = \sqrt{2}$     v) $\tan x = -$ $\frac{1}{\sqrt{3}}$     vi) $\sqrt{3} \sec x = 2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

i) $\sin 2x =$ $\frac{\sqrt{3}}{2}$    ii) $\cos 3x =$ $\frac{1}{2}$    iii) $\sin 9x = \sin x$

iv) $\sin 2x = \cos 3x$    v) $\tan x + \cot 2x = 0$    vi) $\tan 3x = \cot x$

vii) $\tan 2x \tan x = 1$    viii) $\tan mx + \cot nx = 0$     ix) $\tan px = \cot qx$

x) $\sin 2x + \cos x = 0$    xi) $\sin x = \tan x$     xii) $\sin 3x + \cos 2x = 0$

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

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

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

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

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

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

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

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

iii) $\sin 9x = \sin x$

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

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

Therefore, either

 $\cos 5x = 0$                                        or $\Rightarrow \cos 5x = \cos$ $\frac{\pi}{2}$ $\Rightarrow 5x = (2n+1)$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = (2n+1)$ $\frac{\pi}{10}$ $; n \in Z$ $\sin 4x = 0$ $\Rightarrow \sin 4x = \sin 0$ $\Rightarrow 4x = n\pi +(-1)^n ( 0)$ $\Rightarrow x =$ $\frac{n\pi}{4}$ $; n \in Z$

iv) $\sin 2x = \cos 3x$

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

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

 Considering positive sign $3x = 2n\pi + ($ $\frac{\pi}{2}$ $- 2x) ; n \in Z$ $\Rightarrow 5x = ( 4n + 1 )$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = ( 4n + 1 )$ $\frac{\pi}{10}$ $; n \in Z$ Considering negative sign $3x = 2n\pi - ($ $\frac{\pi}{2}$ $- 2x) ; n \in Z$ $\Rightarrow x = ( 4n - 1 )$ $\frac{\pi}{2}$ $; n \in Z$

v) $\tan x + \cot 2x = 0$

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

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

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

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

vi) $\tan 3x = \cot x$

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

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

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

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

vii) $\tan 2x \tan x = 1$

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

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

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

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

viii) $\tan mx + \cot nx = 0$

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

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

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

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

ix) $\tan px = \cot qx$

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

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

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

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

x) $\sin 2x + \cos x = 0$

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

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

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

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

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

 Considering positive sign $x = 2n\pi + ($ $\frac{\pi}{2}$ $+ 2x) ; n \in Z$ $\Rightarrow x = -2n\pi -$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = - ( 4n+1)$ $\frac{\pi}{2}$ $; n \in Z$ Considering negative sign $x = 2n\pi - ($ $\frac{\pi}{2}$ $+ 2x) ; n \in Z$ $\Rightarrow 3x = ( 4n - 1 )$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = (4n-1)$ $\frac{\pi}{6}$ $; n \in Z$

xi) $\sin x = \tan x$

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

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

Therefore, either

 $\sin x = 0$                                          or $\Rightarrow x = n\pi + ( -1)^n (0) ; n \in Z$ $\Rightarrow x = n\pi ; n \in Z$ $( 1 -$ $\frac{1}{\cos x}$ $) = 0$ $\Rightarrow \cos x -1 = 0$ $\Rightarrow x = 2m\pi ; m \in Z$

xii) $\sin 3x + \cos 2x = 0$

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

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

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

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

 Considering positive sign $2x = 2n \pi + ($ $\frac{\pi}{2}$ $+ 3x); n \in Z$ $\Rightarrow - x = ( 4n+1)$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = -( 4n+1)$ $\frac{\pi}{2}$ $; n \in Z$ Considering negative sign $2x = 2n \pi - ($ $\frac{\pi}{2}$ $+ 3x); n \in Z$ $\Rightarrow 5x = ( 4n-1)$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = ( 4n-1)$ $\frac{\pi}{10}$ $; n \in Z$

$\\$

Question 3: Solve the following equations:

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

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

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

vii) $\cos 4x = \cos 2x$

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

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

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

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

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

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

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

Therefore either

 $2 \cos x + 3 = 0$                                or $\Rightarrow \cos x = -$ $\frac{3}{2}$ This is not possible as $-1 \leq \cos x \leq 1$ $2 \cos x -1 = 0$ $\Rightarrow \cos x =$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos$ $\frac{\pi}{3}$ $\Rightarrow x = 2n\pi \pm$ $\frac{\pi}{3}$ $; n \in Z$

ii) $2 \cos^2 x - 5 \cos x + 2 = 0$

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

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

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

Therefore either

 $\cos x - 2 = 0$                                    or $\Rightarrow \cos x = 2$ This is not possible as $-1 \leq \cos x \leq 1$ $2 \cos x -1 = 0$ $\Rightarrow \cos x =$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos$ $\frac{\pi}{3}$ $\Rightarrow x = 2n \pi \pm$ $\frac{\pi}{3}$ $; n \in Z$

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

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

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

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

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

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

Therefore either,

 $\cos x - \sqrt{3} = 0$               or $\Rightarrow \cos x = \sqrt{3}$ $-1 \leq \cos x \leq 1$ $not \ possible$ $2\cos x + \sqrt{3}$ $\Rightarrow \cos x = -$ $\frac{\sqrt{3}}{2}$ $\Rightarrow \cos x = \cos ( \pi -$ $\frac{\pi}{6}$ $)$ $\Rightarrow \cos x = \cos$ $\frac{5\pi}{6}$ $\Rightarrow x = 2n\pi \pm$ $\frac{5\pi}{6}$ $; n \in Z$

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

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

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

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

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

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

Therefore, either

 $2 \cos x + 5 = 0$               or $\Rightarrow \cos x = - 2.5$ $Not \ possible$ $2 \cos x -1 = 0$ $\Rightarrow \cos x =$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos$ $\frac{\pi}{3}$ $\Rightarrow x = 2n \pi \pm$ $\frac{\pi}{3}$ $; n \in Z$

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

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

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

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

Therefore, either

 $\tan x + 1 = 0$                             or $\Rightarrow \tan x = - 1$ $\Rightarrow \tan x = - \tan$ $\frac{\pi}{4}$ $= \tan ( -$ $\frac{\pi}{4}$ $)$ $\Rightarrow x = n\pi -$ $\frac{\pi}{4}$ $; n \in Z$ $\tan x - \sqrt{3} = 0$ $\Rightarrow \tan x = \sqrt{3}$ $\Rightarrow \tan x = \tan$ $\frac{\pi}{3}$ $\Rightarrow x = m\pi +$ $\frac{\pi}{3}$ $; m \in Z$

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

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

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

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

Therefore, either

 $\cos x - \sqrt{3} \sin x = 0$          or $\Rightarrow \tan x =$ $\frac{1}{\sqrt{3}}$ $\Rightarrow \tan x = \tan$ $\frac{\pi}{6}$ $\Rightarrow x = n\pi +$ $\frac{\pi}{6}$ $; n \in Z$ $3 \cos x +\sqrt{3} \sin x = 0$ $\Rightarrow \tan x = - \sqrt{3}$ $\Rightarrow \tan x = \tan ( -$ $\frac{\pi}{3}$ $)$ $\Rightarrow x = m\pi -$ $\frac{\pi}{3}$ $; m \in Z$

vii) $\cos 4x = \cos 2x$

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

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

Therefore, either

 $\sin 3x = 0$                           or $\Rightarrow 3x = n\pi ; n \in Z$ $\sin x = 0$ $\Rightarrow x = m\pi ; m \in Z$

$\\$

Question 4: Solve the following equations:

i) $\cos x + \cos 2x + \cos 3x = 0$    ii) $\cos x + \cos 3x - \cos 2x = 0$

iii) $\sin x + \sin 5x = \sin 3x$     iv) $\cos x \cos 2x \cos 3x =$ $\frac{1}{4}$

v) $\cos x + \sin x = \cos 2x + \sin 2x$      vi) $\sin x + \sin 2x + \sin 3x = 0$

vii) $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$     viii) $\sin 3x - \sin x = 4 \cos^2 x - 2$

ix) $\sin 2x - \sin 4x + \sin 6x=0$

i) $\cos x + \cos 2x + \cos 3x = 0$

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

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

Therefore either

 $\cos 2x = 0$                      or $\Rightarrow \cos 2x = \cos$ $\frac{\pi}{2}$ $\Rightarrow 2x = (2n+1)$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = (2n+1)$ $\frac{\pi}{4}$ $; n \in Z$ $1 + 2 \cos x = 0$ $\Rightarrow \cos x = -$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos ( \pi +$ $\frac{\pi}{3}$ $)$ $\Rightarrow \cos x = \cos$ $\frac{4\pi}{3}$ $\Rightarrow x = m \pi \pm$ $\frac{4\pi}{3}$ $; m \in Z$

ii) $\cos x + \cos 3x - \cos 2x = 0$

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

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

Therefore, either

 $\cos 2x = 0$                        or $\Rightarrow \cos 2x = \cos$ $\frac{\pi}{2}$ $\Rightarrow 2x = (2n+1)$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = (2n+1)$ $\frac{\pi}{4}$ $; n \in Z$ $2 \cos x - 1 = 0$ $\Rightarrow \cos x =$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos$ $\frac{\pi}{3}$ $\Rightarrow x = 2m\pi \pm$ $\frac{\pi}{3}$

iii) $\sin x + \sin 5x = \sin 3x$

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

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

Therefore,either

 $\sin 3x = 0$                                 or $\Rightarrow \sin 3x = \sin 0$ $\Rightarrow 3x = n\pi ; n \in Z$ $\Rightarrow x =$ $\frac{n\pi}{3}$ $; n \in Z$ $2 \cos 2x - 1 = 0$ $\Rightarrow \cos 2x =$ $\frac{1}{2}$ $\Rightarrow \cos 2x = \cos$ $\frac{\pi}{3}$ $\Rightarrow 2x = 2m\pi \pm$ $\frac{\pi}{3}$ $\Rightarrow x = m\pi \pm$ $\frac{\pi}{6}$

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

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

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

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

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

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

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

Therefore, either

 $2 \cos^2 2x - 1 = 0$                       or $\Rightarrow \cos 4x = 0$ $\Rightarrow \cos 4x = \cos$ $\frac{\pi}{2}$ $\Rightarrow 4x = ( 2n+1)$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = ( 2n+1)$ $\frac{\pi}{8}$ $; n \in Z$ $2 \cos 2x + 1 = 0$ $\Rightarrow \cos 2x = -$ $\frac{1}{2}$ $\Rightarrow \cos 2x = \cos$ $\frac{2\pi}{3}$ $\Rightarrow 2x = 2m\pi \pm$ $\frac{2\pi}{3}$ $; n \in Z$ $\Rightarrow x = m\pi \pm$ $\frac{\pi}{3}$ $; n \in Z$

v) $\cos x + \sin x = \cos 2x + \sin 2x$

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

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

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

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

Therefore, either

 $\sin$ $\frac{x}{2}$ $= 0$               or $\Rightarrow$ $\frac{x}{2}$ $= n\pi ; n \in Z$ $\Rightarrow x = 2n\pi ; n \in Z$ $\sin$ $\frac{3x}{2}$ $- \cos$ $\frac{3x}{2}$ $= 0$ $\Rightarrow \tan$ $\frac{3x}{2}$ $= 1$ $\Rightarrow \tan$ $\frac{3x}{2}$ $= \tan$ $\frac{\pi}{4}$ $\Rightarrow$ $\frac{3x}{2}$ $= m\pi +$ $\frac{\pi}{4}$ $\Rightarrow x=$ $\frac{2m\pi}{3}$ $+ \frac{\pi}{6}$ $; m \in Z$

vi) $\sin x + \sin 2x + \sin 3x = 0$

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

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

Therefore, either

 $sin 2x = 0$                             or $\Rightarrow 2x = n\pi ; n \in Z$ $\Rightarrow x =$ $\frac{n\pi}{2}$ $; n \in Z$ $1 + 2 \cos x = 0$ $\Rightarrow \cos x =$ $\frac{-1}{2}$ $= \cos$ $\frac{2\pi}{3}$ $\Rightarrow x = 2m\pi \pm$ $\frac{2\pi}{3}$ $; m \in Z$

vii) $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$

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

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

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

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

Therefore, either

 $\cos x = 0$ $\Rightarrow x = ( 2n + 1 )$ $\frac{\pi}{2}$ $; n \in Z$ or $\sin \frac{5x}{2} = 0$ $\Rightarrow$ $\frac{5x}{2}$ $= m\pi ; m \in Z$ $\Rightarrow x =$ $\frac{2m\pi}{5}$ $; m \in Z$ or $\cos$ $\frac{x}{2}$ $= 0$ $\Rightarrow$ $\frac{x}{2}$ $= ( 2p+1)$ $\frac{\pi}{2}$ $; p \in Z$

viii) $\sin 3x - \sin x = 4 \cos^2 x - 2$

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

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

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

Therefore either,

 $\cos 2x = 0$                           or $\Rightarrow 2x = ( 2n+1)$ $\frac{\pi}{2}$ $; n \in Z$ $\Rightarrow x = ( 2n+1)$ $\frac{\pi}{4}$ $; n \in Z$ $\sin x - 1 = 0$ $\Rightarrow x = m\pi + ( - 1 )^m$ $\frac{\pi}{2}$ $; m \in Z$

ix) $\sin 2x - \sin 4x + \sin 6x=0$

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

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

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

Therefore either,

 $\sin 4x = 0$ $\Rightarrow 4x = n\pi$ $\Rightarrow x =$ $\frac{n\pi}{4}$ $; n \in Z$ $2 \cos 2x - 1 = 0$ $\Rightarrow \cos 2x =$ $\frac{1}{2}$ $\Rightarrow \cos 2x = \cos$ $\frac{\pi}{3}$ $\Rightarrow 2x = 2m\pi \pm$ $\frac{\pi}{3}$ $\Rightarrow x = m\pi \pm$ $\frac{\pi}{6}$

$\\$

Question 5: Solve the following equations:

i) $\tan x + \tan 2x + \tan 3x = 0$    ii) $\tan x + \tan 2x = \tan 3x$

iii) $\tan 3x + \tan x = 2 \tan 2x$

i) $\tan x + \tan 2x + \tan 3x = 0$

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

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

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

Therefore either,

 $\tan x + \tan 2x = 0$ $\Rightarrow \tan x = - \tan 2x$ $\Rightarrow \tan x = \tan ( -2x)$ $\Rightarrow x = n\pi + ( - 2x) ; n \in Z$ $\Rightarrow 3x = n\pi ; n \in Z$ $\Rightarrow x =$ $\frac{n\pi}{3}$ $; n \in Z$ $2 - \tan x \tan 2x = 0$ $\Rightarrow \tan x$ $\frac{2\tan x }{1 - \tan^2 x}$ $= 2$ $\Rightarrow 2 \tan^2 x = 2 - 2 \tan^2 x$ $\Rightarrow 4 \tan^2 x = 2$ $\Rightarrow \tan^2 x =$ $\frac{1}{2}$ $\Rightarrow \tan x = \pm$ $\frac{1}{\sqrt{2}}$ $\Rightarrow x = m\pi \pm \tan^{-1}$ $\frac{1}{\sqrt{2}}$

ii) $\tan x + \tan 2x = \tan 3x$

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

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

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

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

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

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

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

Therefore either,

 $\tan x = 0$ $\Rightarrow x = m\pi ; m \in Z$ or $\tan^2 x = 3$ $\Rightarrow \tan x = \pm \sqrt{3}$ $\Rightarrow \tan x = \pm \tan$ $\frac{\pi}{3}$ $\Rightarrow x = n\pi \pm$ $\frac{\pi}{3}$ $; n \in Z$ or $\tan 2x = 0$ $\Rightarrow 2x = r \pi ; r \in Z$ $\Rightarrow x =$ $\frac{r\pi}{2}$ $; r \in Z$

iii) $\tan 3x + \tan x = 2 \tan 2x$

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

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

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

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

Therefore either,

 $\tan x = 0$ $\Rightarrow x = n\pi ; n \in Z$ or $\tan 2x = 0$ $\Rightarrow 2x = m\pi ; m \in Z$ $\Rightarrow x =$ $\frac{m\pi}{2}$ $; m \in Z$       or $\tan 3x - \tan x = 0$ $\Rightarrow \tan 3x = \tan x$ $\Rightarrow 3x = r\pi + x ; r \in Z$ $\Rightarrow 2x = r \pi ; r \in Z$ $\Rightarrow x =$ $\frac{r\pi}{2}$ $; r \in Z$

$\\$

Question 6: Solve the following equations:

i) $\sin x + \cos x = \sqrt{2}$    ii) $\sqrt{3} \cos x + \sin x = 1$   iii) $\sin x + \cos x = 1$

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

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

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

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

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

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

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

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

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

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

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

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

Therefore either,

 For positive sign $x -$ $\frac{\pi}{6}$ $= 2n\pi +$ $\frac{\pi}{3}$ $\Rightarrow x = 2n\pi +$ $\frac{\pi}{2}$ $; n \in Z$ For negative sign $x -$ $\frac{\pi}{6}$ $= 2n\pi -$ $\frac{\pi}{3}$ $\Rightarrow x = 2n\pi -$ $\frac{\pi}{6}$ $; n \in Z$

iii) $\sin x + \cos x = 1$

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

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

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

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

Therefore either,

 For positive sign $x -$ $\frac{\pi}{4}$ $= 2n\pi +$ $\frac{\pi}{4}$ $\Rightarrow x = 2n\pi +$ $\frac{\pi}{2}$ $; n \in Z$ For negative sign $x -$ $\frac{\pi}{4}$ $= 2n\pi -$ $\frac{\pi}{4}$ $\Rightarrow x = 2n\pi ; n \in Z$

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

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

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

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

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

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

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

Therefore either,

 For positive sign $x -$ $\frac{\pi}{4}$ $= 2n\pi +$ $\frac{\pi}{4}$ $\Rightarrow x = 2n\pi +$ $\frac{\pi}{2}$ $; n \in Z$ For negative sign $x -$ $\frac{\pi}{4}$ $= 2n\pi -$ $\frac{\pi}{4}$ $\Rightarrow x = 2n\pi ; n \in Z$

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

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

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

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

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

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

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

Therefore either,

 For positive sign $x -$ $\frac{5\pi}{12}$ $= 2n\pi +$ $\frac{\pi}{4}$ $\Rightarrow x = 2n\pi +$ $\frac{2\pi}{3}$ $; n \in Z$ For negative sign $x -$ $\frac{5\pi}{12}$ $= 2n\pi -$ $\frac{\pi}{4}$ $\Rightarrow x = 2n\pi +$ $\frac{\pi}{6}$ $; n \in Z$

$\\$

Question 7: Solve the following equations:

i) $\cot x + \tan x = 2$    ii) $2 \sin^2 x = 3 \cos x, 0 \leq x \leq 2\pi$

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

v) $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$

vi) $4 \sin x \cos x + 2 \sin x + 2 \cos x + 1 = 0$    vii) $\cos x + \sin x = \cos 2x + \sin 2x$

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

i) $\cot x + \tan x = 2$

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

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

$\Rightarrow \sin 2x = 1$

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

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

ii) $2 \sin^2 x = 3 \cos x, 0 \leq x \leq 2\pi$

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

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

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

Therefore either,

 $\cos x + 2 = 0$ $\Rightarrow \cos x = - 2$ $Not \ possible$ $2 \cos x -1 = 0$ $\Rightarrow \cos x =$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos$ $\frac{\pi}{3}$ $\Rightarrow x =$ $\frac{\pi}{3}$ $,$ $\frac{5\pi}{3}$

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

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

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

Therefore either,

 $\Rightarrow \cos 3x = 0$ $\Rightarrow 3x =$ $\frac{\pi}{2}$ $\Rightarrow x =$ $\frac{\pi}{6}$ $\Rightarrow \cos 2x = 0$ $\Rightarrow 2x =$ $\frac{\pi}{2}$ $\Rightarrow x =$ $\frac{\pi}{4}$

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

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

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

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

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

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

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

v) $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$

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

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

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

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

Therefore either,

 $2\cos x - 3 = 0$ $\Rightarrow \cos x =$ $\frac{3}{2}$ $Not \ possible$ $\sin 2x - \cos 2x = 0$ $\Rightarrow \tan 2x = 1$ $\Rightarrow \tan 2x = \tan$ $\frac{\pi}{4}$ $\Rightarrow 2x = n\pi +$ $\frac{\pi}{4}$ $\Rightarrow x =$ $\frac{n\pi}{2} +$ $\frac{\pi}{8}$

vi) $4 \sin x \cos x + 2 \sin x + 2 \cos x + 1 = 0$

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

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

Therefore either,

 $2 \cos x + 1 = 0$ $\Rightarrow \cos x = -$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos$ $\frac{2\pi}{3}$ $\Rightarrow x = 2n\pi \pm$ $\frac{2\pi}{3}$ $; n \in Z$ $2 \sin x + 1 = 0$ $\Rightarrow \sin x = -$ $\frac{1}{2}$ $\Rightarrow \sin x = \sin ( \pi +$ $\frac{\pi}{6})$ $\Rightarrow x = m\pi + ( -1)^m$ $\frac{7\pi}{6}$ $; m \in Z$

vii) $\cos x + \sin x = \cos 2x + \sin 2x$

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

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

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

Therefore either,

 $\Rightarrow \sin$ $\frac{x}{2}$ $= 0$ $\Rightarrow$ $\frac{x}{2}$ $= n\pi ; n \in Z$ $\Rightarrow x = 2 n\pi ; n \in Z$ $\Rightarrow \tan$ $\frac{3x}{2}$ $= 1$ $\Rightarrow \tan$ $\frac{3x}{2}$ $= \tan$ $\frac{\pi}{4}$ $\Rightarrow$ $\frac{3x}{2}$ $= m\pi +$ $\frac{\pi}{4}$ $; m \in Z$ $\Rightarrow x =$ $\frac{2m\pi}{3}$ $+$ $\frac{\pi}{6}$ $; m \in Z$

viii) $\sin x \tan x - 1 = \tan x - \sin x$

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

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

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

Therefore either,

 $\sin x - 1 = 0$ $\Rightarrow \sin x = \sin$ $\frac{\pi}{2}$ $\Rightarrow x = n\pi + ( -1 )^n$ $\frac{\pi}{2}$ $\tan x + 1 = 0$ $\Rightarrow \tan x = - 1$ $\Rightarrow \tan x = \tan ($ $\frac{3\pi}{4}$ $)$ $\Rightarrow x = m\pi +$ $\frac{3\pi}{4}$

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

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

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

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

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

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

Therefore either,

 $\Rightarrow \cos x + 3 = 0$ $\Rightarrow \cos x = - 3$ $Not \ possible$ $\Rightarrow 2 \cos x - 1 = 0$ $\Rightarrow \cos x =$ $\frac{1}{2}$ $\Rightarrow \cos x = \cos$ $\frac{\pi}{3}$ $\Rightarrow x = 2n\pi \pm$ $\frac{\pi}{3}$ $; n \in Z$

$\\$

Solve the following equations:

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

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

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

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

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

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

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

Therefore either,

$\sin x - 1 = 0$

$\Rightarrow \sin x = 1$

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

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

or

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

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

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

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

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

With positive sign

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

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

With negative sign

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

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

$\\$

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

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

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

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

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

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

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

Dividing by $\Rightarrow \cos^2 x$

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

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

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

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

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

Therefore either,

 $\Rightarrow \tan x - 3 = 0$        or $\Rightarrow \tan x = 3$ $\Rightarrow x = n\pi + \tan^{-1} 3; n \in Z$ $\Rightarrow \tan x - 2$ $\Rightarrow \tan x = 2$ $\Rightarrow x = m\pi + \tan^{-1} 2 ; m \in Z$

$\\$

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

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

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

if $\Rightarrow 2^{\sin^2 x} = y$

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

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

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

$\Rightarrow y = \sqrt{2}$

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

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

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

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

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