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

Answer:

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

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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

Answer:

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

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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

Answer:

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

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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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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}

Answer:

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