JEE Mathematics Trigonometric Equations MCQs Set A

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Question. The general solution of the equation, \( 2\cos 2x = 3 \cdot 2\cos^2 x - 4 \) is
(a) \( x = 2n\pi \), \( n \in \mathbf{I} \)
(b) \( x = n\pi \), \( n \in \mathbf{I} \)
(c) \( x = n\pi/4 \), \( n \in \mathbf{I} \)
(d) \( x = n\pi/2 \), \( n \in \mathbf{I} \)
Answer: (b) \( x = n\pi \), \( n \in \mathbf{I} \)
Solution:
\( 2 \cos 2x = 3 \cdot 2 \cos^2x - 4 \)
\( \Rightarrow 2 \cos 2x = 3 (\cos 2x + 1) - 4 \)
\( \Rightarrow \cos 2x = 1 \Rightarrow 2x = 2n\pi ; n \in \mathbf{I} \)
\( \Rightarrow x = n\pi ; n \in \mathbf{I} \)

Question. The solution set of the equation \( 4\sin \theta \cdot \cos \theta - 2 \cos \theta - 2\sqrt{3} \sin \theta + \sqrt{3} = 0 \) in the interval \( (0, 2\pi) \) is
(a) \( \left\{ \frac{3\pi}{4}, \frac{7\pi}{4} \right\} \)
(b) \( \left\{ \frac{\pi}{3}, \frac{5\pi}{3} \right\} \)
(c) \( \left\{ \frac{3\pi}{4}, \pi, \frac{\pi}{3}, \frac{5\pi}{3} \right\} \)
(d) \( \left\{ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{11\pi}{6} \right\} \)
Answer: (d) \( \left\{ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{11\pi}{6} \right\} \)
Solution:
\( 4 \sin \theta \cos \theta - 2 \cos \theta - 2\sqrt{3} \sin \theta + \sqrt{3} = 0 \)
\( \Rightarrow 2 \cos \theta (2 \sin \theta - 1) - \sqrt{3} (2 \sin \theta - 1) = 0 \)
\( \Rightarrow (2 \sin \theta - 1) (2 \cos \theta - \sqrt{3}) = 0 \)
\( \Rightarrow \sin \theta = \frac{1}{2} \) or \( \cos \theta = \frac{\sqrt{3}}{2} \)
\( \theta = n\pi + (-1)^n \frac{\pi}{6} \), \( \theta = 2m\pi \pm \frac{\pi}{6} \)
\( \theta = \frac{\pi}{6}, \frac{5\pi}{6} \in (0, 2\pi) \), \( \theta = \frac{\pi}{6}, \frac{11\pi}{6} \)
\( \therefore \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{11\pi}{6} \quad m, n \in \mathbf{I} \)

Question. Total number of solutions of \( \sin x \cdot \tan 4x = \cos x \) belonging to \( (0, \pi) \) are
(a) 4
(b) 7
(c) 8
(d) 5
Answer: (d) 5
Solution:
\( \sin x \tan 4x = \cos x \)
\( \Rightarrow \tan 4x = \cot x \)
\( \Rightarrow \tan 4x = \tan \left(\frac{\pi}{2} - x\right) \)
\( \Rightarrow 4x = n\pi + \frac{\pi}{2} - x \quad n \in \mathbf{I} \)
\( \Rightarrow x = (2n + 1)\frac{\pi}{10} \)
For \( x \in (0, \pi) \quad n \in \mathbf{I} \)
\( (2n + 1) < 10 \)
\( n < \frac{9}{2} \Rightarrow n = 0, 1, 2, 3, 4 \Rightarrow \) No. of sol. is 5

Question. All solutions of the equation, \( 2 \sin \theta + \tan \theta = 0 \) are obtained by taking all integral values of m and n in
(a) \( 2n\pi + \frac{2\pi}{3} \), \( n \in \mathbf{I} \)
(b) \( n\pi \) or \( 2m\pi \pm \frac{2\pi}{3} \) where \( n, m \in \mathbf{I} \)
(c) \( n\pi \) or \( m\pi \pm \frac{\pi}{3} \) where \( n, m \in \mathbf{I} \)
(d) \( n\pi \) or \( 2m\pi \pm \frac{\pi}{3} \) where \( n, m \in \mathbf{I} \)
Answer: (b) \( n\pi \) or \( 2m\pi \pm \frac{2\pi}{3} \) where \( n, m \in \mathbf{I} \)
Solution:
\( 2 \sin \theta + \tan \theta = 0 \quad \cos \theta \neq 0 \)
\( \Rightarrow 2 \sin \theta + \frac{\sin \theta}{\cos \theta} = 0 \)
\( \Rightarrow \sin \theta (2 \cos \theta + 1) = 0 \)
\( \Rightarrow \sin \theta = 0 \) or \( \cos \theta = -\frac{1}{2} \)
\( \theta = n\pi \) or \( \theta = 2m\pi \pm \frac{2\pi}{3} \quad n, m \in \mathbf{I} \)

Question. The most general solution of \( \tan \theta = -1 \) and \( \cos \theta = \frac{1}{\sqrt{2}} \) is
(a) \( n\pi + \frac{7\pi}{4} \), \( n \in \mathbf{I} \)
(b) \( n\pi + (-1)^n \frac{7\pi}{4} \), \( n \in \mathbf{I} \)
(c) \( 2n\pi + \frac{7\pi}{4} \), \( n \in \mathbf{I} \)
(d) None of the options
Answer: (c) \( 2n\pi + \frac{7\pi}{4} \), \( n \in \mathbf{I} \)
Solution:
\( \tan \theta = -1 \) & \( \cos \theta = \frac{1}{\sqrt{2}} \)
principal solution
\( \theta = \frac{3\pi}{4}, \frac{7\pi}{4} \) & \( \theta = \frac{\pi}{4}, \frac{7\pi}{4} \)
common principal solution is \( \frac{7\pi}{4} \)
then general solution is \( \theta = 2n\pi + \frac{7\pi}{4} \), \( n \in \mathbf{I} \)

Question. If \( 2 \cos^2 (\pi + x) + 3 \sin (\pi + x) \) vanishes then the values of x lying in the interval from 0 to \( 2\pi \) are
(a) \( x = \pi/6 \) or \( 5\pi/6 \)
(b) \( x = \pi/3 \) or \( 5\pi/3 \)
(c) \( x = \pi/4 \) or \( 5\pi/4 \)
(d) \( x = \pi/2 \) or \( 5\pi/2 \)
Answer: (a) \( x = \pi/6 \) or \( 5\pi/6 \)
Solution:
\( 2 \cos^2(\pi + x) + 3 \sin(\pi + x) = 0 \)
\( \Rightarrow 2 \cos^2x - 3 \sin x = 0 \)
\( \Rightarrow 2 - 2 \sin^2x - 3 \sin x = 0 \)
\( \Rightarrow (2 \sin x - 1)(\sin x + 2) = 0 \)
\( \Rightarrow \sin x = \frac{1}{2} \) or \( \sin x \neq -2 \)
\( x = \frac{\pi}{6}, \frac{5\pi}{6} \)

Question. If \( 20 \sin^2\theta + 21 \cos \theta - 24 = 0 \) & \( \frac{7\pi}{4} < \theta < 2\pi \) then the values of \( \cot \frac{\theta}{2} \) is
(a) 3
(b) \( \frac{\sqrt{15}}{3} \)
(c) \( -\frac{\sqrt{15}}{3} \)
(d) \( -3 \)
Answer: (d) \( -3 \)
Solution:
\( 20 \sin^2\theta + 21 \cos\theta - 24 = 0 \) & \( \frac{7\pi}{4} < \theta < 2\pi \)
\( \Rightarrow 20 - 20\cos^2\theta + 21 \cos \theta - 24 = 0 \)
\( \Rightarrow 20 \cos^2\theta - 21 \cos\theta + 4 = 0 \)
\( \Rightarrow (5 \cos\theta - 4) (4 \cos \theta - 1) = 0 \)
\( \Rightarrow \cos \theta \neq \frac{1}{4} \quad \because \theta \notin \left(\frac{7\pi}{4}, 2\pi\right) \)
or \( \cos \theta = \frac{4}{5} \quad \Rightarrow \cos \frac{\theta}{2} = \sqrt{\frac{\cos\theta + 1}{2}} \)
\( \cos \frac{\theta}{2} = \pm \frac{3}{\sqrt{10}} \)
\( \cos \frac{\theta}{2} = -\frac{3}{\sqrt{10}} \quad \because \frac{\theta}{2} \in \left(\frac{7\pi}{8}, \pi\right) \)
\( \sin \frac{\theta}{2} = \frac{1}{\sqrt{10}} \Rightarrow \cot \frac{\theta}{2} = -3 \)

Question. If \( x \in \left[0, \frac{\pi}{2}\right] \), the number of solutions of the equation, \( \sin 7x + \sin 4x + \sin x = 0 \) is
(a) 3
(b) 5
(c) 6
(d) None of the options
Answer: (b) 5
Solution:
\( \sin 7x + \sin 4x + \sin x = 0 \)
\( \Rightarrow \sin 7x + \sin x + \sin 4x = 0 \)
\( \Rightarrow 2 \sin 4x \cos 3x + \sin 4x = 0 \)
\( \Rightarrow \sin 4x (2 \cos 3x + 1) = 0 \)
\( \Rightarrow \sin 4x = 0 \Rightarrow 4x = n\pi \quad n \in \mathbf{I}, x \in \left[0, \frac{\pi}{2}\right] \)
\( x = \frac{n\pi}{4} \quad \therefore x = 0, \frac{\pi}{4}, \frac{\pi}{2} \)
or \( \cos 3x = -\frac{1}{2} \)
\( 3x = 2m\pi \pm \frac{2\pi}{3} \Rightarrow x = (6m \pm 2)\frac{\pi}{9} \quad m \in \mathbf{I} \)
\( \therefore x = \frac{2\pi}{9}, \frac{4\pi}{9} \quad \because x \in \left[0, \frac{\pi}{2}\right] \)
No. of sol. 5

Question. The general solution of \( \sin x + \sin 5x = \sin 2x + \sin 4x \) is
(a) \( 2n\pi ; n \in \mathbf{I} \)
(b) \( n\pi ; n \in \mathbf{I} \)
(c) \( n\pi/3 ; n \in \mathbf{I} \)
(d) \( 2n\pi/3 ; n \in \mathbf{I} \)
Answer: (c) \( n\pi/3 ; n \in \mathbf{I} \)
Solution:
\( \sin x + \sin 5x = \sin 2x + \sin 4x \)
\( \Rightarrow 2 \sin 3x \cos 2x = 2 \sin 3x \cos x \)
\( \Rightarrow \sin 3x (\cos 2x - \cos x) = 0 \)
\( \Rightarrow \sin 3x = 0 \Rightarrow 3x = n\pi \Rightarrow x = \frac{n\pi}{3}, n \in \mathbf{I} \)
or \( \cos 2x - \cos x = 0 \)
\( -2 \sin \frac{3x}{2} \sin \frac{x}{2} = 0 \)
or \( \frac{3x}{2} = m\pi \) or \( \frac{x}{2} = k\pi \quad \{m, k \in \mathbf{I}\} \)
\( x = \frac{2m\pi}{3} \) or \( x = 2k\pi \)
Union of all solutions is \( x = \frac{n\pi}{3} \quad n \in \mathbf{I} \)

Question. A triangle ABC is such that \( \sin(2A + B) = \frac{1}{2} \). If A, B, C are in A.P. then the angle A, B, C are respectively
(a) \( \frac{5\pi}{12}, \frac{\pi}{4}, \frac{\pi}{3} \)
(b) \( \frac{\pi}{4}, \frac{\pi}{3}, \frac{5\pi}{12} \)
(c) \( \frac{\pi}{3}, \frac{\pi}{4}, \frac{5\pi}{12} \)
(d) \( \frac{\pi}{3}, \frac{5\pi}{12}, \frac{\pi}{4} \)
Answer: (b) \( \frac{\pi}{4}, \frac{\pi}{3}, \frac{5\pi}{12} \)
Solution:
\( \sin (2A + B) = \frac{1}{2} \)
\( \sin (2A + B) = \sin \frac{\pi}{6} \) or \( \sin \frac{5\pi}{6} \)
\( 2A + B = \frac{\pi}{6} \) is not possible
\( \therefore 2A + B = \frac{5\pi}{6} \) ...(i)
Given that \( 2B = A + C \) ...(ii)
We know \( A + B + C = \pi \) ...(iii)
From (ii) & (iii) \( 3B = \pi \Rightarrow B = \frac{\pi}{3} \)
From (i) \( 2A = \frac{5\pi}{6} - \frac{\pi}{3} \Rightarrow A = \frac{\pi}{4} \)
From (iii) \( C = \pi - (A + B) = \pi - \left(\frac{\pi}{3} + \frac{\pi}{4}\right) \Rightarrow C = \frac{5\pi}{12} \)

Question. \( \frac{\cos 3\theta}{2\cos 2\theta - 1} = \frac{1}{2} \) if
(a) \( \theta = n\pi + \frac{\pi}{3} \), \( n \in \mathbf{I} \)
(b) \( \theta = 2n\pi \pm \frac{\pi}{3} \), \( n \in \mathbf{I} \)
(c) \( \theta = 2n\pi \pm \frac{\pi}{6} \), \( n \in \mathbf{I} \)
(d) \( \theta = n\pi + \frac{\pi}{6} \), \( n \in \mathbf{I} \)
Answer: (b) \( \theta = 2n\pi \pm \frac{\pi}{3} \), \( n \in \mathbf{I} \)
Solution:
\( \frac{\cos 3\theta}{2 \cos 2\theta - 1} = \frac{1}{2} \Rightarrow \frac{4 \cos^3 \theta - 3 \cos \theta}{4 \cos^2 \theta - 2 - 1} = \frac{1}{2} \)
\( \Rightarrow 2 \cos \theta [4 \cos^2 \theta - 3] = (4 \cos^2 \theta - 3) \)
\( \Rightarrow (2 \cos \theta - 1) (4 \cos^2 \theta - 3) = 0 \)
\( \Rightarrow \cos \theta = \frac{1}{2} \Rightarrow \theta = 2n\pi \pm \frac{\pi}{3}, n \in \mathbf{I} \)
or \( \cos^2 \theta = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2 \Rightarrow \theta = n\pi \pm \frac{\pi}{6}, m \in \mathbf{I} \)
but \( \theta = n\pi \pm \frac{\pi}{6} \) doesn't satisfy the given equation.

Question. \( \frac{\sin 3\theta}{2\cos 2\theta + 1} = \frac{1}{2} \) if
(a) \( \theta = n\pi + \frac{\pi}{6} \), \( n \in \mathbf{I} \)
(b) \( \theta = 2n\pi - \frac{\pi}{6} \), \( n \in \mathbf{I} \)
(c) \( \theta = n\pi + (-1)^n \frac{\pi}{6} \), \( n \in \mathbf{I} \)
(d) \( \theta = n\pi - \frac{\pi}{6} \), \( n \in \mathbf{I} \)
Answer: (c) \( \theta = n\pi + (-1)^n \frac{\pi}{6} \), \( n \in \mathbf{I} \)
Solution:
\( \frac{\sin 3\theta}{2 \cos 2\theta + 1} = \frac{1}{2} \)
\( \Rightarrow \frac{3 \sin \theta - 4 \sin^3 \theta}{2 - 4 \sin^2 \theta + 1} = \frac{1}{2} \)
\( \Rightarrow 2 \sin \theta [3 - 4 \sin^2 \theta] = (3 - 4 \sin^2 \theta) \)
\( \Rightarrow (2 \sin \theta - 1) (3 - 4 \sin^2 \theta) = 0 \)
\( \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbf{I} \)
or \( \sin^2 \theta = \frac{3}{4} \Rightarrow \theta = n\pi \pm \frac{\pi}{3}, m \in \mathbf{I} \)
But \( \theta = n\pi \pm \frac{\pi}{3} \) doesn't satisfy the given equation.

Question. If \( \cos 2\theta + 3 \cos \theta = 0 \) then
(a) \( \theta = 2n\pi \pm \alpha \) where \( \alpha = \cos^{-1}\left(\frac{\sqrt{17} - 3}{4}\right) \)
(b) \( \theta = 2n\pi \pm \alpha \) where \( \alpha = \cos^{-1}\left(\frac{-\sqrt{17} - 3}{4}\right) \)
(c) \( \theta = 2n\pi \pm \alpha \) where \( \alpha = \cos^{-1}\left(\frac{\pm \sqrt{17} - 3}{4}\right) \)
(d) None of the options
Answer: (a) \( \theta = 2n\pi \pm \alpha \) where \( \alpha = \cos^{-1}\left(\frac{\sqrt{17} - 3}{4}\right) \)
Solution:
\( \cos 2\theta + 3 \cos \theta = 0 \)
\( \Rightarrow 2 \cos^2\theta + 3 \cos\theta - 1 = 0 \)
\( \Rightarrow \cos\theta = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4} \)
(\(-\)) sign reject
\( \Rightarrow \cos\theta = \cos \alpha \)
\( \theta = 2n\pi \pm \alpha \quad \text{where} \quad \alpha = \cos^{-1}\left(\frac{-3 + \sqrt{17}}{4}\right) \)

Question. If \( \sin \theta + 7 \cos \theta = 5 \), then \( \tan (\theta/2) \) is a root of the equation
(a) \( x^2 - 6x + 1 = 0 \)
(b) \( 6x^2 - x - 1 = 0 \)
(c) \( 6x^2 + x + 1 = 0 \)
(d) \( x^2 - x + 6 = 0 \)
Answer: (b) \( 6x^2 - x - 1 = 0 \)
Solution:
\( \sin \theta + 7 \cos \theta = 5 \)
\( \Rightarrow \frac{2\tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} + 7\left(\frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\right) = 5, \text{ Let } \tan \frac{\theta}{2} = t \)
\( \Rightarrow 2t + 7(1 - t^2) = 5 (1 + t^2) \Rightarrow 6t^2 - t - 1 = 0 \)

Question. The general solution of the equation \( \tan x + \tan \left(x + \frac{\pi}{3}\right) + \tan \left(x + \frac{2\pi}{3}\right) = 3 \) is
(a) \( \frac{n\pi}{4} + \frac{\pi}{12} \), \( n \in \mathbf{I} \)
(b) \( \frac{n\pi}{3} + \frac{\pi}{6} \), \( n \in \mathbf{I} \)
(c) \( \frac{n\pi}{3} + \frac{\pi}{12} \), \( n \in \mathbf{I} \)
(d) None of the options
Answer: (c) \( \frac{n\pi}{3} + \frac{\pi}{12} \), \( n \in \mathbf{I} \)
Solution:
\( \tan x + \tan \left(x + \frac{\pi}{3}\right) + \tan \left(x + \frac{2\pi}{3}\right) = 3 \)
\( \Rightarrow \tan x + \frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x} + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x} = 3 \)
\( \Rightarrow \frac{\tan x - 3 \tan^3 x + \tan x + \sqrt{3} \tan^2 x + \sqrt{3} + 3 \tan x + \tan x - \sqrt{3} \tan^2 x - \sqrt{3} + 3 \tan x}{1 - 3 \tan^2 x} = 3 \)
\( \Rightarrow \frac{9 \tan x - 3 \tan^3 x}{1 - 3 \tan^2 x} = 3 \Rightarrow \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1 \)
\( \Rightarrow \tan 3x = 1 \Rightarrow 3x = n\pi + \frac{\pi}{4} \quad n \in \mathbf{I} \)
\( \Rightarrow x = \frac{n\pi}{3} + \frac{\pi}{12} \quad n \in \mathbf{I} \)

Question. The general solution of the equation \( \tan^2 \alpha + 2\sqrt{3} \tan \alpha = 1 \) is given by
(a) \( \alpha = \frac{n\pi}{2} \), \( n \in \mathbf{I} \)
(b) \( \alpha = (2n + 1) \frac{\pi}{2} \), \( n \in \mathbf{I} \)
(c) \( \alpha = (6n + 1) \frac{\pi}{12} \), \( n \in \mathbf{I} \)
(d) \( \alpha = \frac{n\pi}{12} \), \( n \in \mathbf{I} \)
Answer: (c) \( \alpha = (6n + 1) \frac{\pi}{12} \), \( n \in \mathbf{I} \)
Solution:
\( \tan^2\alpha + 2\sqrt{3} \tan\alpha = 1 \)
\( \Rightarrow 2\sqrt{3} \tan\alpha = 1 - \tan^2\alpha \)
\( \Rightarrow \sqrt{3} \frac{2\tan\alpha}{1 - \tan^2\alpha} = 1 \)
\( \Rightarrow \tan 2\alpha = \frac{1}{\sqrt{3}} = \tan \frac{\pi}{6} \)
\( \Rightarrow 2\alpha = n\pi + \frac{\pi}{6} \quad \because n \in \mathbf{I} \)
\( \Rightarrow \alpha = \frac{n\pi}{2} + \frac{\pi}{12} = (6n + 1)\frac{\pi}{12}, n \in \mathbf{I} \)

Question. \( \sin 3\theta = 4 \sin \theta \cdot \sin 2\theta \cdot \sin 4\theta \) in \( 0 \leq \theta \leq \pi \) has
(a) 2 real solutions
(b) 4 real solutions
(c) 6 real solutions
(d) 8 real solutions
Answer: (d) 8 real solutions
Solution:
\( \sin 3\theta = 4 \sin \theta \cdot \sin 2\theta \sin 4\theta \), \( 0 \leq \theta \leq \pi \)
\( \Rightarrow 3 \sin\theta - 4\sin^3\theta - 4 \sin\theta \sin 2\theta \sin 4\theta = 0 \)
\( \Rightarrow \sin \theta [3 - 4 \sin^2\theta - 4 \sin 2\theta \sin 4\theta] = 0 \)
\( \Rightarrow \sin \theta [3 - 2 + 2 \cos 2\theta - 2 \cos 2\theta + 2 \cos 6\theta] = 0 \)
\( \Rightarrow \sin\theta [1 + 2 \cos 6\theta] = 0 \)
\( \Rightarrow \sin\theta = 0 \) or \( \cos 6\theta = -\frac{1}{2} \)
\( \theta = 0, \pi \) or \( 0 \leq \theta \leq \pi \)
\( 0, \pi \in [0, \pi] \) or \( 0 \leq 6\theta \leq 6\pi \) (3 rounds)
no. of solutions in 1 round = 2
total no. of solutions in \( \theta \in [0, \pi] \) = 2 + 6 = 8

Question. General solution of the equation, \( \cot 3\theta - \cot \theta = 0 \) is
(a) \( \theta = (2n - 1) \frac{\pi}{2} \), \( n \in \mathbf{I} \)
(b) \( \theta = (2n - 1) \frac{\pi}{4} \), \( n \in \mathbf{I} \)
(c) \( \theta = (2n - 1) \frac{\pi}{3} \), \( n \in \mathbf{I} \)
(d) None of the options
Answer: (a) \( \theta = (2n - 1) \frac{\pi}{2} \), \( n \in \mathbf{I} \)
Solution:
\( \cot 3\theta - \cot \theta = 0 \)
\( \Rightarrow \cot 3\theta = \cot \theta \)
\( \Rightarrow 3\theta = n\pi + \theta \)
\( \Rightarrow 2\theta = n\pi \)
\( \Rightarrow \theta = \frac{n\pi}{2} \quad \because n \in \mathbf{I} \)
And \( \theta \neq \frac{n\pi}{3} \) and \( \theta \neq m\pi \)
\( \Rightarrow \theta = (2n - 1), \frac{\pi}{2} \quad n \in \mathbf{I} \)
{\( \cot \theta = 0 \) & \( \cot \theta \neq \infty \), so n = odd int}
\( \Rightarrow \theta = (2n - 1), \frac{\pi}{2} \quad n \in \mathbf{I} \)

Question. The set of values of x for which \( \frac{\tan 3x - \tan 2x}{1 + \tan 3x \tan 2x} = 1 \) is
(a) \( \phi \)
(b) \( \{\pi/4\} \)
(c) \( \{n\pi + \pi/4 \mid n = 1, 2, 3\dots\} \)
(d) \( \{2n\pi + \pi/4 \mid n = 1, 2, 3 \dots\} \)
Answer: (a) \( \phi \)
Solution:
\( \frac{\tan 3x - \tan 2x}{1 + \tan 3x \tan 2x} = 1 \)
\( \Rightarrow \tan (3x - 2x) = 1 \)
\( \Rightarrow \tan x = 1 = \tan \frac{\pi}{4} \)
\( \Rightarrow x = n\pi + \frac{\pi}{4} \quad n \in \mathbf{I} \)
But at this value of x, \( \tan 2x = \infty \)
It's not a solution.
\( \Rightarrow x \in \phi \)

Question. The number of integral values of a for which the equation \( \cos 2x + a \sin x = 2a - 7 \) possesses a solution is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (d) 5
Solution:
\( \cos 2x + a \sin x = 2a - 7 \)
\( \Rightarrow 1 - 2\sin^2x + a \sin x = 2a - 7 \)
\( \Rightarrow 2 \sin^2x - a \sin x + (2a - 8) = 0 \)
\( \Rightarrow \sin x = \frac{a \pm \sqrt{a^2 - 4 \cdot 2(8 - 2a)}}{4} = \frac{a \pm \sqrt{(a - 8)^2}}{4} \)
\( \Rightarrow \sin x = \frac{a \pm (a - 8)}{4} ; -1 \leq \sin x \leq 1 \)
\( \Rightarrow \sin x = \frac{2a - 8}{4} \quad \text{so} \quad \sin x \neq 2 \)
\( \Rightarrow -1 \leq \sin x = \frac{a - 4}{2} \leq 1 \)
\( \Rightarrow -2 \leq a - 4 \leq 2 \)
\( \Rightarrow 2 \leq a \leq 6 \quad \because a \in \mathbf{I} \)
\( a = 2, 3, 4, 5, 6 \) so a has 5 values.

Question. The principal solution set of the equation, \( 2 \cos x = \sqrt{2 + 2\sin 2x} \) is
(a) \( \left\{ \frac{\pi}{8}, \frac{13\pi}{8} \right\} \)
(b) \( \left\{ \frac{\pi}{4}, \frac{13\pi}{8} \right\} \)
(c) \( \left\{ \frac{\pi}{4}, \frac{13\pi}{10} \right\} \)
(d) \( \left\{ \frac{\pi}{8}, \frac{13\pi}{10} \right\} \)
Answer: (a) \( \left\{ \frac{\pi}{8}, \frac{13\pi}{8} \right\} \)
Solution:
\( 2 \cos x = \sqrt{2 + 2\sin 2x} \quad x \in [0, 2\pi] \)
\( \Rightarrow 4 \cos^2x = 2 + 2 \sin 2x \)
\( \Rightarrow 2 \cos^2x - 1 = \sin 2x \)
\( \Rightarrow \cos 2x = \sin 2x \)
\( \Rightarrow \cos 2x = \cos \left(\frac{\pi}{2} - 2x\right) \)
\( \Rightarrow 2x = 2n\pi \pm \left(\frac{\pi}{2} - 2x\right) \)
\( \Rightarrow 2x = 2n\pi + \frac{\pi}{2} - 2x \) & \( 2x = 2n\pi - \frac{\pi}{2} + 2x \)
\( \Rightarrow 4x = \left(\frac{4n + 1}{2}\right)\pi \quad \text{not possible} \)
\( \Rightarrow x = \left(\frac{4n + 1}{8}\right)\pi \)
\( \therefore 0 \leq \left(\frac{4n + 1}{8}\right)\pi \leq 2\pi \)
\( \Rightarrow 0 \leq \frac{4n + 1}{8} \leq 2 \)
\( \Rightarrow -\frac{1}{8} \leq \frac{n}{2} \leq 2 - \frac{1}{8} \Rightarrow -\frac{1}{4} \leq n \leq \frac{15}{4} \)
n = 0, 1, 2, 3
\( x \in [0, 2\pi] \)
\( x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8} \)
Since \( \cos x \ge 0 \), negative cosine values are rejected.
\( x = \frac{\pi}{8}, \frac{13\pi}{8} \)

Question. The number of solution of the equation \( |\sin x| = |\cos 3x| \) in \( [-2\pi, 2\pi] \) is
(a) 32
(b) 28
(c) 24
(d) 30
Answer: (c) 24
Solution:
\( |\sin x| = |\cos 3x| \quad x \in [-2\pi, 2\pi] \)
Case (I) : \( \sin x = \cos 3x \)
\( \cos \left(\frac{\pi}{2} - x\right) = \cos 3x \)
\( 3x = 2n\pi \pm \left(\frac{\pi}{2} - x\right) \quad n \in \mathbf{I} \)
\( 3x = 2n\pi + \frac{\pi}{2} - x \) & \( 3x = 2n\pi - \frac{\pi}{2} + x \)
\( 4x = (4n + 1) \frac{\pi}{2} \quad 2x = (4n - 1) \frac{\pi}{2} \)
\( x = (4n + 1)\frac{\pi}{8} \quad x = (4n - 1) \frac{\pi}{4} \)
\( -2 \leq \frac{4n + 1}{8} \leq 2 \quad -2 \leq \frac{4n - 1}{4} \leq 2 \)
\( -\frac{17}{4} \leq n \leq \frac{15}{4} \quad -\frac{7}{4} \leq n \leq \frac{9}{4} \)
n = –4, –3, –2, –1, 0, 1, 2, 3 \quad (no. of sol. 8)
n = –1, 0, 1, 2 \quad (no. of sol. 4)
Case (II) : \( -\sin x = \cos 3x \)
\( \cos \left(\frac{\pi}{2} + x\right) = \cos 3x \)
\( 3x = 2m\pi \pm \left(\frac{\pi}{2} + x\right) \quad \because m \in \mathbf{I} \)
\( 3x = 2n\pi + \frac{\pi}{2} + x \) & \( 3x = 2n\pi - \frac{\pi}{2} - x \)
\( x = \frac{(4n + 1)\pi}{4} \) & \( x = \frac{(4n - 1)\pi}{8} \)
\( -2 \leq \frac{4n + 1}{4} \leq 2 \quad -2 \leq \frac{4n - 1}{8} \leq 2 \)
\( -\frac{9}{4} \leq n \leq \frac{7}{4} \quad -\frac{15}{4} \leq n \leq \frac{17}{4} \)
n = –2, –1, 0, 1 \quad (no. of sol. 4)
n = –3, –2, –1, 0, 1, 2, 3, 4 \quad (no. of sol. 8)
Total no. of sol. is = 8 + 4 + 4 + 8 = 24

Question. The number of all possible triplets \( (a_1, a_2, a_3) \) such that : \( a_1 + a_2 \cos 2x + a_3 \sin^2x = 0 \) for all x is
(a) 0
(b) 1
(c) 2
(d) infinite
Answer: (d) infinite
Solution:
\( a_1 + a_2 \cos 2x + a_3 \sin^2x = 0 \)
\( a_1 + a_2 (1 - 2\sin^2x) + a_3 \sin^2x = 0 \)
\( (a_1 + a_2) = (2a_2 - a_3) \sin^2x \)
\( \sin^2x = \frac{a_1 + a_2}{2a_2 - a_3} \quad \because 0 \leq \sin^2x \leq 1 \)
\( 0 \leq \frac{a_1 + a_2}{2a_2 - a_3} \leq 1 \)
\( 0 \leq a_1 + a_2 \leq 2a_2 - a_3 \)
\( a_1 + a_2 \geq 0 \quad \dots(\text{i}) \)
\( 2a_2 - a_3 \geq 0 \quad \dots(\text{ii}) \)
\( -a_1 + a_2 - a_3 \geq 0 \quad \dots(\text{iii}) \)
homogenous equation system
\( \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & -1 \\ -1 & 1 & -1 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \geq \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \)
\( |A| = 1(-2 + 1) - 1(0 - 1) + 0 = -1 + 1 = 0 \)
So no. of solution is infinite

Question. The value ‘a’ for which the equation \( 4\text{cosec}^2 (\pi(a + x)) + a^2 - 4a = 0 \) has a real solution is :
(a) \( a = 1 \)
(b) \( a = 2 \)
(c) \( a = 3 \)
(d) None of the options
Answer: (b) \( a = 2 \)
Solution:
\( 4 \text{cosec}^2 (\pi (a + x)) + a^2 - 4a = 0 \)
\( \Rightarrow 4 \text{cosec}^2 (\pi (a + x)) = 4a - a^2 \)
\( \Rightarrow \sin^2 (\pi (a + x)) = \frac{4}{4a - a^2} \)
\( 0 \leq \sin^2 x \leq 1 \)
\( 0 \leq \frac{4}{4a - a^2} \leq 1 \)
\( 0 \leq 4 \leq 4a - a^2 \)
\( 4a - a^2 \geq 4 \)
\( a^2 - 4a + 4 \leq 0 \Rightarrow (a - 2)^2 \leq 0 \)
\( \Rightarrow a = 2 \)

Question. If \( 2 \tan^2x - 5 \sec x - 1 = 0 \) has 7 different roots in \( \left[0, \frac{n\pi}{2}\right] \), \( n \in \mathbf{N} \), then greatest value of n is
(a) 8
(b) 10
(c) 13
(d) 15
Answer: (d) 15

Question. The solution of \( |\cos x| = \cos x - 2\sin x \) is
(a) \( x = n\pi \), \( n \in \mathbf{I} \)
(b) \( x = n\pi + \frac{\pi}{4} \), \( n \in \mathbf{I} \)
(c) \( x = n\pi + (-1)^n \frac{\pi}{4} \), \( n \in \mathbf{I} \)
(d) \( (2n+1)\pi + \frac{\pi}{4} \), \( n \in \mathbf{I} \)
Answer: (d) \( (2n+1)\pi + \frac{\pi}{4} \), \( n \in \mathbf{I} \)
Solution:
\( |\cos x| = \cos x - 2\sin x \)
Case-I : \( \cos x = \cos x - 2 \sin x \) if \( \cos x \geq 0 \)
\( \Rightarrow \sin x = 0 \)
\( \Rightarrow x = n\pi \) but \( \cos x \geq 0 \)
only even integer of \( \pi \)
\( x = 2n\pi \quad n \in \mathbf{I} \)
Case-II :
\( -\cos x = \cos x - 2 \sin x \) if \( \cos x < 0 \)
\( -2 \cos x = -2\sin x \)
\( \Rightarrow \tan x = 1 \)
\( \Rightarrow x = n\pi + \frac{\pi}{4} \quad \in (\text{I \& III quadrant}) \)
But \( \cos x < 0 \) So x will be in III quadrant
s.t. n should be odd integer
\( x = (2m + 1)\pi + \frac{\pi}{4}, m \in \mathbf{I} \)

Question. The number of solutions of \( \sin \theta + 2\sin 2\theta + 3\sin 3\theta + 4\sin 4\theta = 10 \) in \( (0, \pi) \) is
(a) 1
(b) 2
(c) 4
(d) 0
Answer: (d) 0
Solution:
\( \sin\theta + 2\sin 2\theta + 3 \sin 3\theta + 4\sin 4\theta = 10 \) in \( (0, \pi) \)
Using boundary of SM
It's only possible if \( \Rightarrow 1 + 2 + 3 + 4 = 10 \)
\( \sin \theta = 1 \) & \( \sin 2\theta = 1 \) & \( \sin 3\theta = 1 \) & \( \sin 4\theta = 1 \)
\( \theta = 2m\pi + \frac{\pi}{2} \), \( 2\theta = 2n\pi + \frac{\pi}{2} \), \( 3\theta = 2k\pi + \frac{\pi}{2} \), \( 4\theta = 2p\pi + \frac{\pi}{2} \)
\( m \in \mathbf{I} \quad n \in \mathbf{I} \quad k \in \mathbf{I} \quad p \in \mathbf{I} \)
\( \theta = \frac{\pi}{2} \) , \( \theta = \frac{\pi}{4} \) , \( \theta = \frac{\pi}{6}, \frac{5\pi}{6} \) , \( \theta = \frac{\pi}{8}, \frac{5\pi}{8} \)
\( x \in \left\{\frac{\pi}{2}\right\} \cap \left\{\frac{\pi}{4}\right\} \cap \left\{\frac{\pi}{6}, \frac{5\pi}{6}\right\} \cap \left\{\frac{\pi}{8}, \frac{5\pi}{8}\right\} \)
\( x \in \{\} \) no. of solutions is zero.

Question. The arithmetic mean of the roots of the equation \( 4\cos^3x - 4\cos^2x - \cos(\pi + x) - 1 = 0 \) in the interval \( [0, 315] \) is equal to
(a) \( 49\pi \)
(b) \( 50\pi \)
(c) \( 51\pi \)
(d) \( 100\pi \)
Answer: (b) \( 50\pi \)
Solution:
\( 4 \cos^3x - 4 \cos^2x - \cos (\pi + x) - 1 = 0 \)
A.M. of roots \( x \in [0, 315] \)
\( \Rightarrow 4 \cos^2x (\cos x - 1) + (\cos x - 1) = 0 \)
\( \Rightarrow (\cos x - 1) (4 \cos^2x + 1) = 0 \)
\( \Rightarrow \cos x = 1 \) or \( \cos^2 x = -\frac{1}{4} \)
\( x = 2n\pi \quad x \in \mathbf{I} \quad \text{not possible} \)
\( 0 \leq 2n \leq 100 \quad x \in [0, 315] \)
\( 0 \leq n \leq 50 \) or \( [0, 100\pi] \)
\( x = 0, 2\pi, 4\pi, 6\pi, \dots, 100 \pi \)
A.M. \( = \frac{2\pi[0 + 1 + 2 + 3 + \dots + 50]}{51} \)
\( = \frac{2\pi}{51} \cdot \frac{50 \times 51}{2} = 50 \pi \)

Question. The values of x between 0 and \( 2\pi \) which satisfy the equation \( \sin x \cdot \sqrt{8\cos^2 x} = 1 \) are in A.P. The common difference of the A.P. is
(a) \( \frac{\pi}{8} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{3\pi}{8} \)
(d) \( \frac{3\pi}{4} \)
Answer: (b) \( \frac{\pi}{4} \for greatest value of

Question. \(\sin x, \sin 2x, \sin 3x\) are in A.P. if
(a) \(x = n\pi/2, n \in I\)
(b) \(x = n\pi, n \in I\)
(c) \(x = 2n\pi, n \in I\)
(d) \(x = (2n+1)\pi, n \in I\)
Answer: (a) \(x = n\pi/2, n \in I\), (b) \(x = n\pi, n \in I\), (c) \(x = 2n\pi, n \in I\), (d) \(x = (2n+1)\pi, n \in I\)
\(\sin x, \sin 2x, \sin 3x\) in A.P.
\(2 \sin 2x = \sin x + \sin 3x\)
\(\Rightarrow 2 \sin 2x = 2 \sin 2x \cos x \Rightarrow \sin 2x (\cos x - 1) = 0\)
\(\Rightarrow \sin 2x = 0\) or \(\cos x = 1\)
\(2x = n\pi\) or \(x = 2n\pi, \quad n \in I\)
\(x = \frac{n\pi}{2}, n \in I\)
If \(n = 2m \Rightarrow x = m\pi, m \in I\)
If \(m\) is odd \(\Rightarrow x = (2k + 1)\pi, k \in I\)

Question. \(\sin x - \cos^2 x - 1\) assumes the least value for the set of values of \(x\) given by
(a) \(x = n\pi + (-1)^{n + 1} (\pi/6), n \in I\)
(b) \(x = n\pi + (-1)^n (\pi/6), n \in I\)
(c) \(x = n\pi + (-1)^n (\pi/3), n \in I\)
(d) \(x = n\pi - (-1)^n (\pi/6), n \in I\)
Answer: (a) \(x = n\pi + (-1)^{n + 1} (\pi/6), n \in I\), (d) \(x = n\pi - (-1)^n (\pi/6), n \in I\)
\(\sin x - \cos^2 x - 1 = \sin^2 x + \sin x - 2\)
Min value of quadratic expression is \(\frac{-D}{4a}\) at \(\frac{-b}{2a}\) \(\Rightarrow \frac{-D}{4a} = -\frac{1+8}{4} = -\frac{9}{4}\) at \(\frac{-b}{2a} = -\frac{1}{2}\)
\(\Rightarrow\) Given expression has min. value at \(\sin x = -\frac{1}{2}\)
\(\Rightarrow \sin x = \sin\left(-\frac{\pi}{6}\right) \Rightarrow x = n\pi + (-1)^n \left(-\frac{\pi}{6}\right), n \in I\)
\(\Rightarrow x = n\pi + (-1)^{n+1} \frac{\pi}{6}\) or \(n\pi - (-1)^n \frac{\pi}{6}, n \in I\)

Question. \(\sin x + \sin 2x + \sin 3x = 0\) if
(a) \(\sin x = 1/2\)
(b) \(\sin 2x = 0\)
(c) \(\sin 3x = \sqrt{3} /2\)
(d) \(\cos x = - 1/2\)
Answer: (b) \(\sin 2x = 0\), (d) \(\cos x = - 1/2\)
\(\sin x + \sin 2x + \sin 3x = 0\)
\(\Rightarrow \sin 2x + 2 \sin \left(\frac{x + 3x}{2}\right) \cos \left(\frac{3x - x}{2}\right) = 0\)
\(\Rightarrow \sin 2x + 2 \sin 2x \cos x = 0\)
\(\Rightarrow \sin 2x [1 + 2 \cos x] = 0\)
\(\Rightarrow \sin 2x = 0\) or \(\cos x = -\frac{1}{2}\)

Question. \(\cos 4x \cos 8x - \cos 5x \cos 9x = 0\) if
(a) \(\cos 12x = \cos 14 x\)
(b) \(\sin 13 x = 0\)
(c) \(\sin x = 0\)
(d) \(\cos x = 0\)
Answer: (b) \(\sin 13 x = 0\), (c) \(\sin x = 0\)
\(\cos 4x \cdot \cos 8x - \cos 5x \cdot \cos 9x = 0\)
\(\Rightarrow 2 \cos 4x \cos 8x - 2 \cos 5x \cos 9x = 0\)
\(\Rightarrow \cos 12x + \cos 4x - \cos 14 x - \cos 4x = 0\)
\(\Rightarrow \cos 12 x = \cos 14x\)
\(\Rightarrow \cos 12x - \cos 14 x = 0 \Rightarrow 2 \sin x \sin 13x = 0\)
\(\Rightarrow \sin x = 0\) or \(\sin 13 x = 0\)

Question. The general solution of the equation \(\cos x \cdot \cos 6x = - 1\), is
(a) \(x = (2n + 1)\pi, n \in I\)
(b) \(x = 2n\pi, n \in I\)
(c) \(x = (2n - 1)\pi, n \in I\)
(d) None of the options
Answer: (a) \(x = (2n + 1)\pi, n \in I\), (c) \(x = (2n - 1)\pi, n \in I\)
\(\cos x \cos 6x = - 1\)
\(\Rightarrow \cos 7x + \cos 5x = -2\)
It's only possible that
\(\cos 7x = -1\) & \(\cos 5x = -1\)
\(7x = (2n + 1) \pi\) & \(5x = (2m + 1)\pi, n, m \in I\)
\(x = (2n + 1) \frac{\pi}{7}\) & \(x = (2m + 1) \frac{\pi}{5}\)
\(x = \frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}\)
& \(x = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5}\)
common solution in one round is \(\pi\)
In general \(x = (2n \pm 1) \pi ,n \in I\)

Question. If \(\sin(x - y) = \cos (x + y) = 1/2\) then the values of \(x\) & \(y\) lying between \(0\) and \(\pi\) are given by
(a) \(x = \pi/4, y = 3\pi/4\)
(b) \(x = \pi/4, y = \pi/12\)
(c) \(x = 5\pi/4, y = 5\pi/12\)
(d) \(x = 11\pi/12, y = 3\pi/4\)
Answer: (b) \(x = \pi/4, y = \pi/12\), (c) \(x = 5\pi/4, y = 5\pi/12\), (d) \(x = 11\pi/12, y = 3\pi/4\)
If \(\sin (x - y) = \cos (x + y) = \frac{1}{2} \quad 0 < x, y < \pi\)
\(\sin (x - y) = \frac{1}{2}\) & \(\cos (x + y) = \frac{1}{2}\)
\(x - y = \frac{\pi}{6}, \frac{5\pi}{6}, (x + y) = \frac{\pi}{3}, \left(2\pi - \frac{\pi}{3}\right)\)
\(x + y = \frac{\pi}{3}, \frac{5\pi}{3}\)
{add small & subtraction is large not possible}
\(x - y \neq \frac{5\pi}{6}\) & \(x + y \neq \frac{\pi}{3}\)
So \(x - y = \frac{\pi}{6}\) or \(x - y = \frac{\pi}{6}\) or \(x - y = \frac{5\pi}{6}\)
& \(x + y = \frac{\pi}{3} \quad x + y = \frac{5\pi}{3} \quad x+y = \frac{5\pi}{3}\)
\(x = \frac{\pi}{4}, y= \frac{\pi}{12} \quad x= \frac{11\pi}{12}, y= \frac{3\pi}{4} \quad x= \frac{5\pi}{4}, y= \frac{5\pi}{12}\)

Question. The equation \(2 \sin \frac{x}{2} \cdot \cos^2x + \sin^2x = 2 \sin \frac{x}{2} \cdot \sin^2x + \cos^2x\) has a root for which
(a) \(\sin 2x = 1\)
(b) \(\sin 2x = -1\)
(c) \(\cos x = \frac{1}{2}\)
(d) \(\cos 2x = -\frac{1}{2}\)
Answer: (a) \(\sin 2x = 1\), (b) \(\sin 2x = -1\), (c) \(\cos x = \frac{1}{2}\), (d) \(\cos 2x = -\frac{1}{2}\)
\(2 \sin \frac{x}{2} \cdot \cos^2x + \sin^2x = 2 \sin \frac{x}{2} \cdot \sin^2x + \cos^2x\)
\(\Rightarrow 2 \sin \frac{x}{2} (\cos^2x - \sin^2x) - (\cos^2x - \sin^2x) = 0\)
\(\Rightarrow (\cos^2x - \sin^2x) (2 \sin \frac{x}{2} - 1) = 0\)
\(\Rightarrow \cos 2x (2 \sin \frac{x}{2} - 1) = 0\)
\(\Rightarrow \cos 2x = 0\) or \(\sin \frac{x}{2} = \frac{1}{2}\)
\(\Rightarrow \sin 2x = \pm 1\) or \(\cos^2 \frac{x}{2} = 1 - \frac{1}{4} = \frac{3}{4}\)
\(\Rightarrow \cos x = 2 \left(\frac{3}{4}\right) - 1 = \frac{1}{2}\)
\(\Rightarrow \cos 2x = 2 \left(\frac{1}{4}\right) - 1 = -\frac{1}{2}\)

Question. \(\cos 15 x = \sin 5x\) if
(a) \(x = -\frac{\pi}{20} + \frac{n\pi}{5}, n \in I\)
(b) \(x = \frac{\pi}{40} + \frac{n\pi}{10}, n \in I\)
(c) \(x = \frac{3\pi}{20} + \frac{n\pi}{5}, n \in I\)
(d) \(x = -\frac{3\pi}{40} + \frac{n\pi}{10}, n \in I\)
Answer: (a) \(x = -\frac{\pi}{20} + \frac{n\pi}{5}, n \in I\), (b) \(x = \frac{\pi}{40} + \frac{n\pi}{10}, n \in I\), (c) \(x = \frac{3\pi}{20} + \frac{n\pi}{5}, n \in I\), (d) \(x = -\frac{3\pi}{40} + \frac{n\pi}{10}, n \in I\)
\(\cos 15x = \sin 5x\) if
\(\cos 15x = \cos \left(\frac{\pi}{2} - 5x\right)\)
\(15x = 2n\pi \pm \left(\frac{\pi}{2} - 5x\right)\)
\(15x = 2n\pi + \frac{\pi}{2} - 5x\) or \(15x = 2n\pi - \frac{\pi}{2} + 5x\)
\(20x = 2n\pi + \frac{\pi}{2}, n \in I\) or \(10 x = 2n\pi - \frac{\pi}{2}, n \in I\)
\(x = \frac{n\pi}{10} + \frac{\pi}{40}, n \in I\) or \(x = \frac{n\pi}{5} - \frac{\pi}{20}, n \in I\)
Put \(n = n - 1 \quad \quad \quad \quad \text{put } n = n + 1\)
\(x = \frac{n\pi}{10} - \frac{\pi}{10} + \frac{\pi}{40} \quad \quad x = \frac{n\pi}{5} + \frac{\pi}{5} - \frac{\pi}{20}\)
\(x = \frac{n\pi}{10} - \frac{3\pi}{40}, n \in I \quad \quad x = \frac{n\pi}{5} + \frac{3\pi}{20}, n \in I\)

Question. \(5 \sin^2x + \sqrt{3} \sin x \cos x + 6 \cos^2x = 5\) if
(a) \(\tan x = -1/\sqrt{3}\)
(b) \(\sin x = 0\)
(c) \(x = n\pi + \pi/2, n \in I\)
(d) \(x = n\pi + \pi/6, n \in I\)
Answer: (a) \(\tan x = -1/\sqrt{3}\), (c) \(x = n\pi + \pi/2, n \in I\)
\(5 \sin^2x + \sqrt{3} \sin x \cos x + 6\cos^2x = 5(1)^2\)
\(\Rightarrow 5 \sin^2x + \sqrt{3} \sin x \cos x + 6 \cos^2x = 5 \sin^2x + 5 \cos^2x\)
\(\Rightarrow \cos^2x + \sqrt{3} \sin x \cos x = 0\)
\(\Rightarrow \cos x [\cos x + \sqrt{3} \sin x] = 0\)
\(\Rightarrow \cos x \left[\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x\right] = 0\)
\(\Rightarrow \cos x \left[\cos \left(x - \frac{\pi}{3}\right)\right] = 0\)
\(\Rightarrow \cos x = 0\) or \(\cos \left(x - \frac{\pi}{3}\right) = 0\)
\(\Rightarrow x = (2n + 1) \frac{\pi}{2}\) or \(x - \frac{\pi}{3} = 2m\pi \pm \frac{\pi}{2}; m \in I\)
\(\Rightarrow x = n\pi + \frac{\pi}{2}, n \in I\) or \(x = 2m\pi + \frac{\pi}{2} + \frac{\pi}{3}; m \in I\)
\(\Rightarrow x = 2m\pi - \frac{\pi}{2} + \frac{\pi}{3}\) or \(x = 2m\pi - \frac{\pi}{6}\)
\(\tan x = -\frac{1}{\sqrt{3}}\)

Question. \(\sin^2 x + 2 \sin x \cos x - 3 \cos^2 x = 0\) if
(a) \(\tan x = 3\)
(b) \(\tan x = -1\)
(c) \(x = n\pi + \pi/4, n \in I\)
(d) \(x = n\pi + \tan^{-1} (-3), n \in I\)
Answer: (c) \(x = n\pi + \pi/4, n \in I\), (d) \(x = n\pi + \tan^{-1} (-3), n \in I\)
\(\sin^2x + 2 \sin x \cos x - 3\cos^2x = 0\)
\(\Rightarrow \sin^2x + 3 \sin x \cos x - \sin x \cos x - 3 \cos^2x = 0\)
\(\Rightarrow (\sin x + 3 \cos x) (\sin x - \cos x) = 0\)
\(\Rightarrow \sin x + 3 \cos x = 0\) or \(\sin x - \cos x = 0\)
\(\because \sin x \neq 0\) & \(\cos x \neq 0\)
\(\Rightarrow \tan x = -3\) or \(\tan x = 1\)
\(x = n\pi + \tan^{-1}(-3)\) or \(x = n\pi + \frac{\pi}{4}; n \in I\)

Question. \(\sin^2x - \cos 2x = 2 - \sin 2x\) if
(a) \(x = n\pi/2, n \in I\)
(b) \(\tan x = 3/2\)
(c) \(x = (2n + 1) \pi/2, n \in I\)
(d) \(x = n\pi + (-1)^n \sin^{-1} (2/3), n \in I\)
Answer: (b) \(\tan x = 3/2\), (c) \(x = (2n + 1) \pi/2, n \in I\)
\(\sin^2x - \cos 2x = 2 - \sin 2x\)
\(\Rightarrow \sin^2x - 2\cos^2x + 1 = 2 - \sin 2x\)
\(\Rightarrow 1 - \cos^2x - 2 \cos^2x + 1 = 2 - \sin 2x\)
\(\Rightarrow \sin 2x - 3\cos^2x = 0\)
\(\Rightarrow \cos x [2\sin x - 3 \cos x] = 0\)
\(\Rightarrow \cos x = 0 \quad \text{or} \quad 2 \sin x - 3 \cos x = 0\)
\(\Rightarrow x = (2n + 1) \frac{\pi}{2}, n \in I\) or \(2 \sin x = 3 \cos x\)
\(\tan x = \frac{3}{2}\)