Trigonometric Equations JEE Mathematics Worksheets Set 01

Subjective Questions

Question. What are the most general values of \( \theta \) which satisfy the equations,
(a) \( \sin \theta = \frac{1}{\sqrt{2}} \)
(b) \( \tan(x - 1) = \sqrt{3} \)
(c) \( \tan \theta = -1 \)
(d) \( \operatorname{cosec} \theta = \frac{2}{\sqrt{3}} \)
(e) \( 2\cot^2 \theta = \operatorname{cosec}^2 \theta \)
Answer:
(a) \( \sin \theta = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} \)
\( \Rightarrow \theta = n\pi + (-1)^n \frac{\pi}{4}, n \in I \)
(b) \( \tan(x - 1) = \sqrt{3} = \tan \frac{\pi}{3} \)
\( (x - 1) = n\pi + \frac{\pi}{3}, n \in I \)
\( x = n\pi + \frac{\pi}{3} + 1, n \in I \)
(c) \( \tan \theta = -1 = \tan\left(-\frac{\pi}{4}\right) \)
\( \theta = n\pi + \left(-\frac{\pi}{4}\right) = n\pi - \frac{\pi}{4}, n \in I \)
(d) \( \operatorname{cosec} \theta = \frac{2}{\sqrt{3}} \)
\( \Rightarrow \sin \theta = \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3} \)
\( \Rightarrow \theta = n\pi + (-1)^n \frac{\pi}{3}, n \in I \)
(e) \( 2\cot^2 \theta = \operatorname{cosec}^2 \theta \)
\( \Rightarrow \cot^2 \theta = \operatorname{cosec}^2 \theta - \cot^2 \theta \)
\( \Rightarrow \cot^2 \theta = 1 = \cot^2 \frac{\pi}{4} \)
\( \Rightarrow \theta = n\pi \pm \frac{\pi}{4}; n \in I \)

Question. Solve : \( \sin 9\theta = \sin \theta \)
Answer: \( \sin 9\theta = \sin \theta \)
\( 9\theta = 2n\pi + \theta \) or \( 9\theta = (2n + 1)\pi - \theta, n \in I \)
\( \Rightarrow 8\theta = 2n\pi \) or \( 10\theta = (2n + 1)\pi, n \in I \)
\( \Rightarrow \theta = \frac{n\pi}{4} \) or \( \theta = (2n + 1)\frac{\pi}{10}; n \in I \)

Question. Solve : \( \cot \theta + \tan \theta = 2 \operatorname{cosec} \theta \)
Answer: \( \cot \theta + \tan \theta = 2 \operatorname{cosec} \theta \)
\( \Rightarrow \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{2}{\sin \theta} \)
\( \Rightarrow \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{2}{\sin \theta} \)
If \( \sin \theta \neq 0 \) & \( \cos \theta \neq 0 \)
\( \Rightarrow \sin \theta - 2\sin \theta \cos \theta = 0 \)
\( \Rightarrow \sin \theta (1 - 2\cos \theta) = 0 \)
\( \Rightarrow \) but \( \sin \theta \neq 0 \) so \( \cos \theta = \frac{1}{2} = \cos \frac{\pi}{3} \)
\( \theta = 2n\pi \pm \frac{\pi}{3}; n \in I \)

Question. Solve : \( \sin 2\theta = \cos 3\theta \)
Answer: \( \sin 2\theta = \cos 3\theta \)
\( \Rightarrow \cos\left(\frac{\pi}{2} - 2\theta\right) = \cos 3\theta \)
\( \Rightarrow 3\theta = 2n\pi \pm \left(\frac{\pi}{2} - 2\theta\right) \)
\( \Rightarrow 5\theta = 2n\pi + \frac{\pi}{2} \) or \( \theta = 2n\pi - \frac{\pi}{2} \)
\( \Rightarrow \theta = \frac{2n\pi}{5} + \frac{\pi}{10} \) or \( \theta = 2n\pi - \frac{\pi}{2}, n \in I \)
\( = \left(2n + \frac{1}{2}\right)\frac{\pi}{5}, n \in I \)

Question. Solve : \( \cot \theta = \tan 8\theta \)
Answer: \( \cot \theta = \tan 8\theta \)
\( \Rightarrow \tan\left(\frac{\pi}{2} - \theta\right) = \tan 8\theta \Rightarrow 8\theta = n\pi + \frac{\pi}{2} - \theta \)
\( \Rightarrow 9\theta = n\pi + \frac{\pi}{2} \Rightarrow \theta = \frac{n\pi}{9} + \frac{\pi}{18}; n \in I \)
\( \Rightarrow \theta = \left(n + \frac{1}{2}\right)\frac{\pi}{9}; n \in I \)

Question. Solve : \( \tan^2 \theta - (1 + \sqrt{3})\tan \theta + \sqrt{3} = 0 \)
Answer: \( \tan^2 \theta - (1 + \sqrt{3})\tan \theta + \sqrt{3} = 0 \)
\( (\tan \theta - 1)(\tan \theta - \sqrt{3}) = 0 \)
\( \tan \theta = 1 \) or \( \tan \theta = \sqrt{3} \)
\( \theta = n\pi + \frac{\pi}{4}, n \in I \) or \( \theta = n\pi + \frac{\pi}{3}; n \in I \)

Question. Find all the angles between \( 0^\circ \) and \( 90^\circ \) which satisfy the equation \( \sec^2 \theta \cdot \operatorname{cosec}^2 \theta + 2 \operatorname{cosec}^2 \theta = 8 \)
Answer: \( \theta \in (0^\circ, 90^\circ) \)
\( \sec^2 \theta \cdot \operatorname{cosec}^2 \theta + 2 \operatorname{cosec}^2 \theta = 8 \)
\( \Rightarrow \frac{1 + 2\cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = 8 \quad \) diff \( \sin \theta \neq 0 \) , \( \cos \theta \neq 0 \)
\( \Rightarrow 1 + 2\cos^2 \theta = 8(1 - \cos^2 \theta)\cos^2 \theta \)
\( \Rightarrow 8\cos^4 \theta - 6\cos^2 \theta + 1 = 0 \)
\( \Rightarrow (4\cos^2 \theta - 1)(2\cos^2 \theta - 1) = 0 \)
\( \Rightarrow \cos^2 \theta = \left(\frac{1}{2}\right)^2 \) or \( \cos^2 \theta = \left(\frac{1}{\sqrt{2}}\right)^2 \)
\( \Rightarrow \cos \theta = \frac{1}{2} \) or \( \cos \theta = \frac{1}{\sqrt{2}} \)
\( \Rightarrow \theta = \frac{\pi}{3} \) or \( \theta = \frac{\pi}{4} \)

Question. Solve : \( 4\cos \theta - 3\sec \theta = 2\tan \theta \)
Answer: \( 4\cos^2 \theta - 3 = 2\sin \theta \ (\because \cos \theta \neq 0) \)
\( \Rightarrow 4 - 4\sin^2 \theta - 3 = 2\sin \theta \)
\( \Rightarrow 4\sin^2 \theta + 2\sin \theta - 1 = 0 \)
\( \Rightarrow \sin \theta = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-1 \pm \sqrt{5}}{4} \)
\( \sin \theta = \frac{\sqrt{5} - 1}{4} \) or \( \sin \theta = -\frac{(\sqrt{5} + 1)}{4} \)
\( \Rightarrow \sin \theta = \sin \frac{\pi}{10} \) or \( \sin \theta = -\cos\left(\frac{\pi}{5}\right) \)
\( \Rightarrow \theta = n\pi + (-1)^n \frac{\pi}{10} \) or \( \sin \theta = \sin\left(\frac{-3\pi}{10}\right) \)
\( \theta = m\pi + (-1)^n \left(\frac{-3\pi}{10}\right), m \in I \)

Question. Solve : \( \cot \theta - \tan \theta = 2 \)
Answer: \( \cot \theta - \tan \theta = 2 \)
\( 1 - \tan^2 \theta = 2\tan \theta; \quad \tan \theta \neq 0 \)
\( \tan^2 \theta + 2\tan \theta - 1 = 0 \)
\( \tan \theta = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2} \)
\( \tan \theta = \sqrt{2} - 1 \) or \( \tan \theta = -(\sqrt{2} + 1) \)
\( \tan \theta = \tan \frac{\pi}{8} \quad \tan \theta = -\tan \frac{3\pi}{8} \)
\( \theta = n\pi + \frac{\pi}{8}, n \in I \) or \( \tan \theta = \tan\left(\frac{-3\pi}{8}\right) \)
\( \theta = \left(2n + \frac{1}{4}\right)\frac{\pi}{2}, n \in I \) or \( \theta = m\pi + \frac{-3\pi}{8} \)
\( \theta = m\pi - \frac{\pi}{2} + \frac{\pi}{8} \)
\( \theta = (2m - 1)\frac{\pi}{2} + \frac{\pi}{8} \)
\( \theta = \left((2m - 1) + \frac{1}{4}\right)\frac{\pi}{2}, m \in I \)
combined \( \theta = \left(n + \frac{1}{4}\right)\frac{\pi}{2}, n \in I \)

Question. Solve : \( \sin \theta + \sin 3\theta + \sin 5\theta = 0 \)
Answer: \( \sin \theta + \sin 3\theta + \sin 5\theta = 0 \)
\( \sin 3\theta + 2\sin 3\theta \cos 2\theta = 0 \)
\( \sin 3\theta(1 + 2\cos 2\theta) = 0 \)
\( \sin 3\theta = 0 \) or \( 1 + 2\cos 2\theta = 0 \)
\( 3\theta = n\pi \quad \cos 2\theta = -\frac{1}{2} = \cos\left(\frac{2\pi}{3}\right) \)
\( \theta = \frac{n\pi}{3}, n \in I \quad 2\theta = 2m\pi \pm \left(\frac{2\pi}{3}\right) \)
\( \theta = m\pi \pm \frac{\pi}{3}, m \in I \)

Question. Solve : \( \cos \theta + \sin \theta = \cos 2\theta + \sin 2\theta \)
Answer: \( \cos \theta + \sin \theta = \cos 2\theta + \sin 2\theta \)
\( \Rightarrow \frac{1}{\sqrt{2}}(\cos \theta + \sin \theta) = \frac{1}{\sqrt{2}}(\cos 2\theta + \sin 2\theta) \)
\( \Rightarrow \cos\left(\theta - \frac{\pi}{4}\right) = \cos\left(2\theta - \frac{\pi}{4}\right) \)
\( \Rightarrow 2n\pi \pm \left(\theta - \frac{\pi}{4}\right) = 2\theta - \frac{\pi}{4} \)
\( \Rightarrow 2\theta - \frac{\pi}{4} = 2n\pi + \theta - \frac{\pi}{4} \Rightarrow \theta = 2n\pi, n \in I \)
& \( 2\theta - \frac{\pi}{4} = 2n\pi - \theta + \frac{\pi}{4} \)
\( \Rightarrow 3\theta = 2n\pi + \frac{\pi}{2} \Rightarrow \theta = \frac{2n\pi}{3} + \frac{\pi}{6}, n \in I \)

Question. Find all values of \( \theta \) between \( 0^\circ \) & \( 180^\circ \) satisfying the equation; \( \cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0 \)
Answer: \( \theta \in (0, \pi) \)
\( \cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0 \)
\( 2\cos 5\theta \cos \theta + 2\cos^2 \theta = 0 \)
\( 2\cos \theta [\cos 5\theta + \cos \theta] = 0 \)
\( \cos \theta = 0 \) or \( \cos 2\theta = 0 \) or \( \cos 3\theta = 0 \)
\( \theta = \frac{\pi}{2} \) or \( \theta = \frac{\pi}{4} \) or \( 3\theta = (2k + 1)\frac{\pi}{2} \)
\( \theta = \frac{\pi}{4}, \frac{3\pi}{4} \) or \( \theta = \frac{\pi}{6}, \frac{5\pi}{6} \)
\( \theta = \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{6} \)

Question. Solve : \( \sin^2 n\theta - \sin^2(n - 1)\theta = \sin^2 \theta \), where n is constant and \( n \neq 0, 1 \)
Answer: \( \sin^2 n\theta - \sin^2(n - 1)\theta = \sin^2 \theta \), n is constant & \( n \neq 0, 1 \)
\( \Rightarrow \sin(n\theta + n\theta - \theta)\sin(n\theta - n\theta + \theta) = \sin^2 \theta \)
\( \Rightarrow \sin(2n\theta - \theta)\sin \theta = \sin^2 \theta \)
\( \Rightarrow \sin \theta = 0 \) or \( \sin(2n - 1)\theta - \sin \theta = 0 \)
\( \theta = m\pi, m \in I \) or \( 2\cos\left(\frac{2n\theta - \theta + \theta}{2}\right)\sin\left(\frac{2n\theta - \theta - \theta}{2}\right) = 0 \)
\( 2\cos(n\theta)\sin(n - 1)\theta = 0 \)
\( \cos n\theta = 0 \) or \( \sin(n - 1)\theta = 0 \)
\( n\theta = (2k + 1)\frac{\pi}{2} \) or \( (n - 1)\theta = p\pi \quad k, p \in I \)
\( \theta = \frac{(2k + 1)\pi}{2n} \) or \( \theta = \frac{p\pi}{(n - 1)}; p \in I \)

Question. Solve : \( \sqrt{3}\sin \theta - \cos \theta = \sqrt{2} \)
Answer: \( \sqrt{3}\sin \theta - \cos \theta = \sqrt{2} \)
\( \Rightarrow \frac{\sqrt{3}}{2}\sin \theta - \frac{1}{2}\cos \theta = \frac{1}{\sqrt{2}} \)
\( \Rightarrow \sin\left(\theta - \frac{\pi}{6}\right) = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} \)
\( \Rightarrow \theta - \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{4} \)
\( \Rightarrow \theta = n\pi + \frac{\pi}{6} + (-1)^n \frac{\pi}{4}; n \in I \)

Question. Solve : \( \operatorname{cosec} \theta = \cot \theta + \sqrt{3} \)
Answer: \( \operatorname{cosec} \theta = \cot \theta + \sqrt{3} \)
\( \Rightarrow \operatorname{cosec}^2 \theta = \cot^2 \theta + 2\sqrt{3}\cot \theta + 3 \)
\( \Rightarrow \operatorname{cosec}^2 \theta - \cot^2 \theta = 2\sqrt{3}\cot \theta + 3 \)
\( \Rightarrow 1 = 2\sqrt{3}\cot \theta + 3 \Rightarrow \cot \theta = -\frac{2}{2\sqrt{3}} \)
\( \Rightarrow \tan \theta = -\sqrt{3} \Rightarrow \tan \theta = \tan\left(-\frac{\pi}{3}\right) \)
\( \theta = n\pi + \left(-\frac{\pi}{3}\right) \)
\( \theta = n\pi - \frac{\pi}{3}, n \in I \)
but IV quadrant is reject
Aliter :
\( \frac{1}{\sin \theta} = \frac{\cos \theta}{\sin \theta} + \sqrt{3}, \sin \theta \neq 0 \)
\( \sqrt{3}\sin \theta + \cos \theta = 1 \)
\( \frac{\sqrt{3}}{2}\sin \theta + \frac{1}{2}\cos \theta = \frac{1}{2} \)
\( \cos\left(\theta - \frac{\pi}{3}\right) = \frac{1}{2} \)
\( \theta - \frac{\pi}{3} = 2m\pi \pm \frac{\pi}{3}, m \in I \)
\( \theta = 2m\pi \) or \( \theta = 2m\pi + \frac{2\pi}{3} \quad m \in I \)
reject \( \operatorname{cosec} \theta \) & \( \cot \theta \) is present. because does not satisfy given equation (in IV quadrant)

Question. Solve : \( 5\sin \theta + 2\cos \theta = 5 \)
Answer: \( 5\sin \theta + 2\cos \theta = 5 \)
\( \Rightarrow \frac{10\tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} + \frac{2(1 - \tan^2 \frac{\theta}{2})}{1 + \tan^2 \frac{\theta}{2}} = 5 \)
\( \Rightarrow 10\tan \frac{\theta}{2} + 2 - 2\tan^2 \frac{\theta}{2} = 5 + 5\tan^2 \frac{\theta}{2} \)
\( \Rightarrow 7\tan^2 \frac{\theta}{2} - 10\tan \frac{\theta}{2} + 3 = 0 \)
\( \Rightarrow \left(\tan \frac{\theta}{2} - 1\right)\left(7\tan \frac{\theta}{2} - 3\right) = 0 \)
\( \Rightarrow \frac{\theta}{2} = n\pi + \frac{\pi}{4} \) or \( \frac{\theta}{2} = n\pi + \alpha \)
\( \Rightarrow \theta = 2n\pi + \frac{\pi}{2}, n \in I \)
or \( \theta = 2n\pi + 2\alpha; \alpha =

Advanced Subjective Questions

Question. Solve : \( \sin 5x = \cos 2x \) for all values of \( x \) between \( 0^\circ \) & \( 180^\circ \).
Answer:  \( \sin 5x = \cos 2x \)
\( \Rightarrow \sin 5x - \sin \left( \frac{\pi}{2} - 2x \right) = 0 \)
\( \Rightarrow \cos \left( \frac{3x + \frac{\pi}{2}}{2} \right) \sin \left( \frac{7x - \frac{\pi}{2}}{2} \right) = 0 \)
\( \Rightarrow \cos \left( \frac{3x}{2} + \frac{\pi}{4} \right) = 0 \text{ or } \sin \left( \frac{7x}{2} - \frac{\pi}{4} \right) = 0 \)
\( \frac{3x}{2} + \frac{\pi}{4} = (2n + 1)\frac{\pi}{2}, n \in I \text{ or } \frac{7x}{2} - \frac{\pi}{4} = n\pi, n \in I \)
\( 3x = 2n\pi + \frac{\pi}{2}, n \in I \text{ or } 7x = 2n\pi + \frac{\pi}{2} \)
\( x = (4n + 1)\frac{\pi}{6}, n \in I \text{ or } x = (4n + 1)\frac{\pi}{14} \)
\( x = 30^\circ, 150^\circ \quad x = \frac{90^\circ}{7}, \frac{450^\circ}{7}, \frac{810^\circ}{7}, \frac{1170^\circ}{7} \)

Question. Find the solution set of the equation,
\( \log_{\frac{-x^2 - 6x}{10}} (\sin 3x + \sin x) = \log_{\frac{-x^2 - 6x}{10}} (\sin 2x) \).
Answer:  \( \log_{\frac{-x^2 - 6x}{10}} (\sin 3x + \sin x) = \log_{\frac{-x^2 - 6x}{10}} (\sin 2x) \)
\( \Rightarrow \sin 3x + \sin x = \sin 2x \)
\( \Rightarrow 2 \sin 2x \cos x - \sin 2x = 0 \)
\( \Rightarrow 2 \cos x - 1 = 0 \quad \because \sin 2x > 0 \)
\( \Rightarrow \cos x = \frac{1}{2} \Rightarrow x = 2n\pi \pm \frac{\pi}{3}, n \in I \quad \dots(i) \)
\( \because \frac{-x^2 - 6x}{10} > 0 \Rightarrow x(x + 6) < 0 \Rightarrow x \in (-6, 0) \)
so by (i) , \( x = \left\{ \frac{-5\pi}{3}, \frac{-4\pi}{3}, -\pi, \frac{-2\pi}{3}, \frac{-\pi}{3} \right\} \)
but \( \sin 2x > 0 \Rightarrow x = \left\{ -\frac{5\pi}{3}, -\frac{2\pi}{3} \right\} \)
& \( \sin 3x + \sin x > 0 \Rightarrow 3 \sin x - 4\sin^3 x + \sin x > 0 \)
\( \Rightarrow 4 \sin x (1 - \sin^2 x) > 0 \Rightarrow 4 \sin x \cos^2 x > 0 \)
\( \Rightarrow \sin x > 0 \) Only when \( x = \frac{-5\pi}{3} \)
\( \sin x < 0 \) if \( x = \frac{-2\pi}{3} \quad \therefore x = -\frac{5\pi}{3} \)

Question. Find the value of \( \theta \), which satisfy
\( 3 - 2 \cos \theta - 4 \sin \theta - \cos 2\theta + \sin 2\theta = 0 \).
Answer: \( 3 - 2 \cos \theta - 4 \sin \theta - \cos 2\theta + \sin 2\theta = 0 \)
\( \Rightarrow 2 - 2 \cos \theta - 2 \sin \theta + \sin 2\theta - 2 \sin \theta + (1 - \cos 2\theta) = 0 \)
\( \Rightarrow 2(1 - \cos \theta)(1 - \sin \theta) - 2 \sin \theta (1 - \sin \theta) = 0 \)
\( \Rightarrow (1 - \sin \theta) [1 - \cos \theta - \sin \theta] = 0 \)
\( \Rightarrow \sin \theta = 1 \text{ or } \sin \theta + \cos \theta = 1 \)
\( \theta = 2n\pi + \frac{\pi}{2}, n \in I \text{ or } \cos\left(\theta - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
or \( \theta - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4} \)
or \( \theta = 2n\pi + \frac{\pi}{2} \text{ or } \theta = 2n\pi \)
\( \therefore x \in \theta = 2n\pi, 2n\pi + \frac{\pi}{2}, n \in I \)

Question. Find the general solution of the equation,
\( \sin \pi x + \cos \pi x = 0 \). Also find the sum of all solutions in \( [0, 100] \).
Answer:  \( \sin \pi x + \cos \pi x = 0 \quad \because x \in [0, 100] \)
\( \Rightarrow \cos\left(\pi x - \frac{\pi}{4}\right) = 0 \)
\( \pi x - \frac{\pi}{4} = (2n + 1)\frac{\pi}{2} \)
\( x = n + \frac{1}{2} + \frac{1}{4} \)
\( x = n + \frac{3}{4}, n \in I \text{ or } n - \frac{1}{4}, n \in I \)
sum \( = \left(0 + \frac{3}{4}\right) + \left(1 + \frac{3}{4}\right) + \left(2 + \frac{3}{4}\right) + \dots + \left(99 + \frac{3}{4}\right) \)
\( = (1 + 2 + 3 + 99) + \frac{3}{4} \times 100 \)
\( = \frac{99 \times 100}{2} + 75 = 4950 + 75 = 5025 \)

Question. Find the range of \( y \) such that the equation , \( y + \cos x = \sin x \) has a real solution . For \( y = 1 \), find \( x \) such that \( 0 < x < 2\pi \).
Answer: \( y + \cos x = \sin x \)
\( \Rightarrow \sin x - \cos x = y \Rightarrow -\sqrt{2} \leq \sin x - \cos x \leq \sqrt{2} \)
\( \Rightarrow -\sqrt{2} \leq y \leq \sqrt{2} \)
\( y = 1 \quad \sin\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)
\( x - \frac{\pi}{4} = \frac{\pi}{4}, \frac{3\pi}{4} \Rightarrow x = \frac{\pi}{2}, \pi \)

Question. Find the general values of \( \theta \) for which the quadratic function \( (\sin \theta) x^2 + (2 \cos \theta) x + \frac{\cos \theta + \sin \theta}{2} \) is the square of a linear function.
Answer: \( (\sin \theta) x^2 + (2 \cos \theta) x + \left( \frac{\cos \theta + \sin \theta}{2} \right) \)
\( b^2 - 4ac = 0 \)
\( 4 \cos^2 \theta - 4 \sin \theta \left( \frac{\cos \theta + \sin \theta}{2} \right) = 0 \)
\( \Rightarrow 2\cos^2 \theta - \sin \theta (\cos \theta + \sin \theta) = 0 \)
\( \Rightarrow 2 \cos^2 \theta - \sin \theta \cos \theta - \sin^2 \theta = 0 \)
\( \Rightarrow 2 \cos \theta (\cos \theta - \sin \theta) + \sin \theta (\cos \theta - \sin \theta) = 0 \)
\( \Rightarrow (\cos \theta - \sin \theta) (2 \cos \theta + \sin \theta) = 0 \)
\( \Rightarrow \cos \theta = \cos\left(\frac{\pi}{2} - \theta\right) \text{ or } \frac{2}{\sqrt{5}} \cos \theta + \frac{1}{\sqrt{5}} \sin \theta = 0 \)
\( \theta = 2n\pi \pm \left(\frac{\pi}{2} - \theta\right) \quad \sin(\theta + \alpha) = 0 \text{ } \{\tan \alpha = 2\} \)
\( \theta = 2n\pi \pm \frac{\pi}{2} - \theta \quad \theta + \alpha = 2n\pi \)
\( \theta = n\pi \pm \frac{\pi}{4} \)
\( \sin \theta = \cos \theta = \pm \frac{1}{\sqrt{2}} \)
\( f(\theta) = \pm \frac{1}{\sqrt{2}} x^2 + \left(\pm \frac{2}{\sqrt{2}}\right) x + \left(\pm \frac{1}{\sqrt{2}}\right) \)
\( = \pm \frac{1}{\sqrt{2}} (x + 1)^2 \)
with \( (-) \) sign not in square of linear
\( = + \frac{1}{\sqrt{2}} (x + 1)^2 \quad \{ q^1 (2n + 1)x + \frac{\pi}{4} \} \)
\( = \left( \frac{x + 1}{2^{1/4}} \right)^2 \)
\( \theta = n\pi + \frac{\pi}{4} \)
\( \theta = 2n\pi - \tan^{-1} 2 \)
\( \tan(\pi - \alpha) = \tan \alpha = 2 \)
\( \pi - \alpha = \tan^{-1} 2 \)
\( \alpha = \pi - \tan^{-1} 2 \)
\( \theta = (2n + 1)\pi - \tan^{-1} 2 \)

Question. Find the general solution of the equation,
\( \tan^2(x + y) + \cot^2(x + y) = 1 - 2x - x^2 \).
Answer: \( \tan^2 (x + y) + \cot^2 (x + y) = 1 - 2x - x^2 \)
L.H.S. \( = \tan^2 (x + y) + \frac{1}{\tan^2(x + y)} \)
min value is = 2
L.H.S. \( \geq 2 \)
R.H.S. = max. value \( = -\frac{D}{4a} = \frac{-(4 + 4)}{4(-1)} = 2 \)
R.H.S. \( \leq 2 \)
\( \therefore \text{L.H.S.} = \text{R.H.S.} = 2 \)
\( \therefore \tan^2 (x + y) + \cot^2 (x + y) = 2 \)
\( \Rightarrow \tan (x + y) = \cot (x + y) \)
\( \Rightarrow \tan^2 (x + y) = 1 + \tan^2 \left(\frac{\pi}{4}\right) \)
\( \Rightarrow (x + y) = n\pi \pm \frac{\pi}{4} \Rightarrow y = n\pi \pm \frac{\pi}{4} + 1, n \in I \)
& \( 1 - 2x - x^2 = 2 \Rightarrow x^2 + 2x + 1 = 0 \)
\( \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1 \)

Question. Prove that the equations
(a) \( \sin x \cdot \sin 2x \cdot \sin 3x = 1 \)
(b) \( \sin x \cdot \cos 4x \cdot \sin 5x = -1/2 \)
have no solution
Answer: \( \sin x \cdot \sin 2x \cdot \sin 3x = 1 \)
Let \( 0 < \sin x < 1 \Rightarrow \sin 2x \sin 3x > 1 \)
But it is not possible
\( \Rightarrow \sin x = 1 \text{ \& } \sin 2x = 1 \text{ \& } \sin 3x = 1 \)
But It's not possible \( \Rightarrow \) no sol.

Question. Find the general solution of \( \sec 4\theta - \sec 2\theta = 2 \).
Answer: \( \sec 4\theta - \sec 2\theta = 2 \)
\( \Rightarrow \cos 2\theta - \cos 4\theta = 2 \cos 2\theta \cos 4\theta \)
\( \Rightarrow \cos 2\theta - \cos 4\theta = \cos 6\theta + \cos 2\theta \)
\( \Rightarrow \cos 6\theta + \cos 4\theta = 0 \)
\( \Rightarrow 2 \cos 5\theta \cos \theta = 0 \)
\( \Rightarrow \cos 5\theta = 0 \text{ or } \cos \theta = 0 \)
\( 5\theta = (2n + 1)\frac{\pi}{2}, n \in I \quad \theta = (2x + 1)\frac{\pi}{2}, n \in I \)
or \( 5\theta = 2n\pi \pm \frac{\pi}{2}, n \in I \text{ or } \theta = 2n\pi \pm \frac{\pi}{2}, n \in I \)
\( \theta = \frac{2n\pi}{5} \pm \frac{\pi}{10}, n \in I \text{ or } \theta = 2n\pi \pm \frac{\pi}{2}, n \in I \)
\( \therefore \sec 4\theta \text{ \& } \sec 2\theta \text{ defined at that angles.} \)

Question. Solve the equation : \( \frac{\sqrt{3}}{2} \sin x - \cos x = \cos^2 x \).
Answer: \( \frac{\sqrt{3}}{2} \sin x - \cos x = \cos^2 x \)
\( \Rightarrow \sqrt{3} \sin x = 2 (\cos x + \cos^2 x) \)
\( \Rightarrow 3 \sin^2 x = 4 (\cos^2 x + \cos^4 x + 2 \cos^3 x) \)
squaring
\( \Rightarrow 3 - 3\cos^2 x = 4 \cos^2 x + 4 \cos^4 x + 8 \cos^3 x \)
\( \Rightarrow \text{Let } \cos x = t \)
\( \Rightarrow 4t^4 + 8t^3 + 7t^2 - 3 = 0 \)
\( \Rightarrow (t + 1) (4t^3 + 4t^2 + 3t - 3) = 0 \)
\( \Rightarrow t = -1 \text{ or } t = \frac{1}{2} \text{ or } 2t^2 + 3t + 3 \neq 0 \quad \because D < 0 \)
\( \Rightarrow \cos x = -1 \text{ or } \cos x = \frac{1}{2} \)
\( x = 2n\pi \pm \pi, n \in I \text{ or } x = 2n\pi \pm \frac{\pi}{3}, n \in I \)

Question. If \( \alpha \) & \( \beta \) are the roots of the equation,
\( a \cos \theta + b \sin \theta = c \) then prove that :
(i) \( \sin \alpha + \sin \beta = \frac{2bc}{a^2 + b^2} \)
(ii) \( \sin \alpha \cdot \sin \beta = \frac{c^2 - a^2}{a^2 + b^2} \)
(iii) \( \tan \frac{\alpha}{2} + \tan \frac{\beta}{2} = \frac{2b}{a + c} \)
(iv) \( \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2} = \frac{c - a}{c + a} \)
Answer: \( a \cos \theta + b \sin \theta = c \)
\( \Rightarrow a \cos \theta = c - b\sin \theta \) squaring
\( \Rightarrow a^2\cos^2 \theta = c^2 + b^2 \sin^2 \theta - 2bc \sin \theta \)
\( \Rightarrow a^2 - a^2\sin^2 \theta = c^2 + b^2 \sin^2 \theta - 2bc \sin \theta \)
\( \Rightarrow (a^2 + b^2) \sin^2 \theta - 2bc \sin \theta + (c^2 - a^2) = 0 \)
(i) \( \sin \alpha + \sin \beta = \frac{2bc}{a^2 + b^2} \)
(ii) \( \sin \alpha \sin \beta = \frac{c^2 - a^2}{a^2 + b^2} \)
\( a \cos \theta + b \sin \theta = c \)
\( \Rightarrow a \left( \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \right) + b \left( \frac{2\tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \right) = c \)
\( \Rightarrow a - a\tan^2 \frac{\theta}{2} + 2b \tan \frac{\theta}{2} = c + c \tan^2 \frac{\theta}{2} \)
\( \Rightarrow (a + c) \tan^2 \frac{\theta}{2} - 2b \tan \frac{\theta}{2} + (c - a) = 0 \)
(iii) \( \tan \frac{\alpha}{2} + \tan \frac{\beta}{2} = \frac{2b}{a + c} \)
(iv) \( \tan \frac{\alpha}{2} \tan \frac{\beta}{2} = \frac{c - a}{c + a} \)

Question. Find the general solution of the following equation : \( 2(\sin x - \cos 2x) - \sin 2x(1+2\sin x)+2 \cos x = 0 \).
Answer:  \( 2(\sin x - \cos 2x) - \sin 2x (1+2 \sin x)+2 \cos x = 0 \)
\( \Rightarrow 2 (\sin x - \cos 2x) - 2 \sin x \cos x (1 + 2\sin x) + 2 \cos x = 0 \)
\( \Rightarrow 2 (\sin x - \cos 2x) - 2 \cos x [\sin x + (2 \sin^2 x - 1)] = 0 \)
\( \Rightarrow 2(\sin x - \cos 2x) - 2 \cos x [\sin x - \cos 2x] = 0 \)
\( \Rightarrow 2 (\sin x - \cos 2x) (1 - \cos x) = 0 \)
\( \Rightarrow \cos x = 1 \Rightarrow x = 2n\pi, n \in I \)
or \( \sin x - \cos 2x = 0 \)
\( 2\sin^2 x +\sin x - 1 = 0 \)
\( (\sin x + 1) (2\sin x - 1) = 0 \)
\( \sin x = -1 \text{ or } \sin x = \frac{1}{2} \)
\( x = n\pi + (-1)^n \left(-\frac{\pi}{2}\right), n \in I \)
or \( x = n\pi + (-1)^n \left(\frac{\pi}{6}\right) n \in I \)

Question. Solve the inequality \( \sin 2x > \sqrt{2} \sin^2 x + (2 - \sqrt{2}) \cos^2 x \).
Answer: \( \sin 2x > \sqrt{2} \sin^2 x + (2 - \sqrt{2}) \cos^2 x \)
\( 2\sin x \cos x > \sqrt{2} \sin^2 x + (2 - \sqrt{2}) \cos^2 x \)
divided by \( \sqrt{2} \cos^2 x \quad \because \cos^2 x > 0 \)
{ \( \because \cos x \neq 0 \), If \( \cos x = 0 \Rightarrow 0 > \sqrt{2} \sin^2 x \) which is false.}
\( \therefore \sqrt{2} \tan x > \tan^2 x + (\sqrt{2} - 1) \)
\( \Rightarrow \tan^2 x - \sqrt{2} \tan x + (\sqrt{2} - 1) < 0 \)
\( \Rightarrow (\tan x - 1) [\tan x - (\sqrt{2} - 1)] < 0 \)
\( (\sqrt{2} - 1) < \tan x < 1 \)
\( \Rightarrow n\pi + \frac{\pi}{8} < x < n\pi + \frac{\pi}{4} \quad n \in I \)

Question. Find the values of \( x \), between \( 0 \) & \( 2\pi \), satisfying the equation ;
\( \cos 3x + \cos 2x = \sin \frac{3x}{2} + \sin \frac{x}{2} \).
Answer: \( \cos 3x + \cos 2x = \sin \frac{3x}{2} + \sin \frac{x}{2}, x \in (0, 2\pi) \)
\( \Rightarrow 2 \cos \frac{5x}{2} \cdot \cos \frac{x}{2} = 2 \sin x \cos \frac{x}{2} \)
\( \Rightarrow 2\cos \frac{x}{2} \left[ \cos \frac{5x}{2} - \sin x \right] = 0 \)
\( \Rightarrow \cos \frac{x}{2} = 0 \text{ or } \cos \frac{5x}{2} = \cos \left(\frac{\pi}{2} - x\right) \)
\( \Rightarrow \frac{x}{2} = \frac{\pi}{2}, \frac{3\pi}{2} \quad \therefore x = \pi \)
\( \frac{5x}{2} = 2n\pi + \frac{\pi}{2} - x \text{ or } \frac{5x}{2} = 2n\pi - \frac{\pi}{2} + x \)
\( \frac{7x}{2} = (4n + 1)\frac{\pi}{2} \text{ or } \frac{3x}{2} = (4n - 1)\frac{\pi}{2} \)
\( x = (4n + 1)\frac{\pi}{7} \quad x = (4n - 1)\frac{\pi}{3} \)
\( x = \frac{\pi}{7}, \frac{5\pi}{7}, \frac{9\pi}{7}, \frac{13\pi}{7} \quad x = -\frac{\pi}{3}, \pi, \frac{7\pi}{3} \)
\( \frac{7\pi}{3} > 2\pi \)

Question. Find the set of values of x satisfying the equality
\( \sin \left(x - \frac{\pi}{4}\right) - \cos \left(x + \frac{3\pi}{4}\right) = 1 \)
and the inequality \( \frac{2 \cos 7x}{\cos 3 + \sin 3} > 2^{\cos 2x} \).
Answer: \( \sin \left(x - \frac{\pi}{4}\right) - \cos \left(x + \frac{3\pi}{4}\right) = 1 \)
\( \Rightarrow \sin \left(x - \frac{\pi}{4}\right) - \cos \left(\frac{\pi}{2} + x + \frac{\pi}{4}\right) = 1 \)
\( \Rightarrow \sin \left(x - \frac{\pi}{4}\right) + \sin \left(x + \frac{\pi}{4}\right) = 1 \)
\( \Rightarrow 2 \sin x \cos \frac{\pi}{4} = 1 \Rightarrow \frac{2}{\sqrt{2}} \sin x = 1 \)
\( \Rightarrow \sin x = \frac{1}{\sqrt{2}} \)
\( x = 2n\pi + \frac{\pi}{4} \text{ or } x = 2n\pi + \frac{3\pi}{4} \)
Check \( x = \frac{\pi}{4} \)
\( \Rightarrow \frac{2\cos 7x}{\cos 3 + \sin 3} > 2^{\cos 2x} \Rightarrow \frac{2 \frac{1}{\sqrt{2}}}{(-ve)} > 2^\circ \text{ wrong} \)
Check \( x = \frac{3\pi}{4} \)
\( \Rightarrow \frac{2\cos \frac{21\pi}{4}}{(-ve)} > 2^{\cos \frac{6\pi}{4}} \Rightarrow \frac{2 \left(-\frac{1}{\sqrt{2}}\right)}{(-ve)} > 2^\circ \text{ True} \)
\( \therefore x = 2n\pi + \frac{3\pi}{4}, n \in I \)

Question. Find the sum of all the roots of the equation , \( \sin \sqrt{x} = -1 \), which are less than \( 100\pi^2 \). Also Find the sum of the square roots of these roots . Now , can we conclude that all the roots \( \cos \sqrt{x} = 0 \) are also the roots of \( \sin \sqrt{x} = -1 \)? Justify your answer .
Answer: \( \sin \sqrt{x} = -1 \quad x > 0 \)
\( = -\sin \frac{\pi}{2} = \sin \frac{3\pi}{2} \quad x \neq -\frac{\pi}{2} \)
\( \sqrt{x} = 2n\pi + \frac{3\pi}{2} \quad n \in I \)
\( \sqrt{x} = \frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2}, \frac{15\pi}{2}, \frac{19\pi}{2}, \frac{23\pi}{2}, \dots \)
\( x = \frac{9\pi^2}{4}, \frac{49\pi^2}{4}, \frac{121\pi^2}{4}, \frac{225\pi^2}{4}, \frac{361\pi^2}{4}, \dots, \frac{529\pi^2}{4}, \dots \)
Sum of root which are less than \( 100\pi^2 \)
\( = \frac{9\pi^2}{4} + \frac{49\pi^2}{4} + \frac{121\pi^2}{4} + \frac{225\pi^2}{4} + \frac{361\pi^2}{4} = \frac{765\pi^2}{4} \)
Sum of square root of these roots
\( = \frac{3\pi}{2} + \frac{7\pi}{2} + \frac{11\pi}{2} + \frac{15\pi}{2} + \frac{19\pi}{2} = \frac{55\pi}{2} \)
Now, \( \cos \sqrt{x} = 0 \quad x > 0 \)
\( \Rightarrow \cos \sqrt{x} = \cos \frac{\pi}{2} \)
\( \Rightarrow \sqrt{x} = 2n\pi + \frac{\pi}{2} \text{ or } \sqrt{x} = 2n\pi + \frac{3\pi}{2} \)
\( \Rightarrow x = \left(2n\pi + \frac{\pi}{2}\right)^2 \quad x = \left(2n\pi + \frac{3\pi}{2}\right)^2 \)
So we can conclude the all the roots of \( \cos \sqrt{x} = 0 \) are not also the roots of \( \sin \sqrt{x} = -1 \)

Question. Solve : \( \sin \left(\frac{\sqrt{x}}{2}\right) + \cos \left(\frac{\sqrt{x}}{2}\right) = \sqrt{2} \sin \sqrt{x} \).
Answer: \( \sin \left(\frac{\sqrt{x}}{2}\right) + \cos \left(\frac{\sqrt{x}}{2}\right) = \sqrt{2} \sin \sqrt{x} \quad x > 0 \)
\( \Rightarrow \frac{1}{\sqrt{2}} \sin \left(\frac{\sqrt{x}}{2}\right) + \frac{1}{\sqrt{2}} \cos \left(\frac{\sqrt{x}}{2}\right) = \sin \sqrt{x} \)
\( \Rightarrow \sin \left(\frac{\sqrt{x}}{2} + \frac{\pi}{4}\right) = \sin \sqrt{x} \)
\( \Rightarrow \frac{\sqrt{x}}{2} + \frac{\pi}{4} = 2n\pi + \sqrt{x} \)
\( \Rightarrow -\frac{\sqrt{x}}{2} = 2n\pi - \frac{\pi}{4} \)
\( \Rightarrow x = \left(4n\pi - \frac{\pi}{2}\right)^2, n \in I \)
or \( \frac{\sqrt{x}}{2} + \frac{\pi}{4} = 2n\pi + \pi - \sqrt{x} \)
\( \Rightarrow \frac{3\sqrt{x}}{2} = 2n\pi + \frac{3\pi}{4} \Rightarrow \sqrt{x} = \frac{4n\pi}{3} + \frac{\pi}{2} \)
\( \Rightarrow x = \left(\frac{4n\pi}{3} + \frac{\pi}{2}\right)^2 \quad n \in I \)

Question. Find the general solution of the equation,
\( \sin \frac{2x + 1}{x} + \sin \frac{2x + 1}{3x} - 3 \cos^2 \frac{2x + 1}{3x} = 0 \)
Answer: \( \sin \left(\frac{2x + 1}{x}\right) + \sin \left(\frac{2x + 1}{3x}\right) - 3 \cos^2 \left(\frac{2x + 1}{3x}\right) = 0 \)
\( \Rightarrow \left\{ \sin \theta + \sin \frac{\theta}{3} \right\} - 3\cos^2 \frac{\theta}{3} = 0 \quad (\text{Let } \frac{2x + 1}{x} = \theta) \)
\( \Rightarrow 2 \sin \frac{2\theta}{3} \cos \frac{\theta}{3} - 3\cos^2 \frac{\theta}{3} = 0 \)
\( \Rightarrow 2 \cdot 2 \sin \frac{\theta}{3} \cos^2 \frac{\theta}{3} - 3\cos^2 \frac{\theta}{3} = 0 \)
\( \Rightarrow \cos^2 \frac{\theta}{3} \left[ 4\sin \frac{\theta}{3} - 3 \right] = 0 \Rightarrow \cos^2 \left(\frac{2x + 1}{3x}\right) = 0 \)
\( \Rightarrow \frac{2x + 1}{3x} = (2n + 1)\frac{\pi}{2} \Rightarrow 4x + 2 = 6n\pi x + 3\pi x \)
\( \Rightarrow x = \frac{2}{(6n\pi + 3\pi - 4)}, n \in I \)
or \( \sin \left(\frac{2x + 1}{3x}\right) = \frac{3}{4} = \sin \alpha \)
\( \Rightarrow \frac{2x + 1}{3x} = n\pi + (-1)^n \alpha \)
\( \Rightarrow 2x + 1 = 3n\pi x + 3(-1)^n \alpha x \)
\( \Rightarrow x = \frac{1}{3n\pi - 2 + 3(-1)^n \sin^{-1} \frac{3}{4}}, n \in I \)

Question. Solve the equation : \( \sin 5x = 16 \sin^5 x \).
Answer: \( \sin 5x = 16\sin^5 x \)
\( \Rightarrow \sin 5x = \sin(2x + 3x) \)
\( = \sin 2x \cos 3x + \cos 2x \sin 3x \)
\( = 2 \sin x \cos x (4\cos^3 x - 3\cos x) + (1 - 2\sin^2 x) (3\sin x - 4\sin^3 x) \)
\( = 2 \sin x (1 - \sin^2 x) (4\cos^2 x - 3) + \sin x (1 - 2 \sin^2 x) (3 - 4\sin^2 x) \)
\( = 2 \sin x (1 - \sin^2 x) (1 - 4\sin^2 x) + \sin x (1 - 2\sin^2 x) (3 - 4 \sin^2 x) \)
\( = 2\sin x - 10\sin^3 x + 8\sin^5 x + 3\sin x - 10\sin^3 x + 8 \sin^5 x \)
\( = 16\sin^5 x - 20 \sin^3 x + 5\sin x \)
\( \Rightarrow \sin 5x = 16 \sin^5 x \Rightarrow - 20\sin^3 x + 5\sin x = 0 \)
\( \Rightarrow 5\sin x [1 - 4\sin^2 x] = 0 \)
\( \Rightarrow \sin x = 0 \text{ or } \sin^2 x = \left(\frac{1}{2}\right)^2 \)
\( \Rightarrow x = n\pi, n \in I \text{ or } x = n\pi \pm \frac{\pi}{6}, n \in I \)

Question. Solve for x & y :
\( x \cos^3 y + 3x \cos y \sin^2 y = 14 \)
\( x \sin^3 y + 3x \cos^2 y \sin y = 13 \)
Answer: \( x \cos^3 y + 3x \cos y \sin^2 y = 14 \quad \dots(i) \)
\( x \sin^3 y + 3x \cos^2 y \sin y = 13 \quad \dots(ii) \)
divide
\( \Rightarrow \frac{\cos^3 y + 3\cos y \sin^2 y}{\sin^3 y + 3\cos^2 y \sin y} = \frac{14}{13} \)
Applying components & dividendo
\( \Rightarrow \frac{(\cos y + \sin y)^3}{(\cos y - \sin y)^3} = \frac{14 + 13}{14 - 13} \)
\( \Rightarrow \left(\frac{\cos y + \sin y}{\cos y - \sin y}\right)^3 = 27 = (3)^3 \)
\( \Rightarrow \frac{\cos y + \sin y}{\cos y - \sin y} = 3 \Rightarrow \frac{1 + \tan y}{1 - \tan y} = 3 \)
\( \Rightarrow \tan y = \frac{1}{2} = \tan \alpha \)
\( \Rightarrow y = n\pi + \tan^{-1} \left(\frac{1}{2}\right), n \in I \)
\( \tan y = \frac{1}{2} \Rightarrow \sin y = \pm \frac{1}{\sqrt{5}}, \cos y = \pm \frac{2}{\sqrt{5}} \)
Case- I: y in Ist quadrant
\( \therefore \sin y = \frac{1}{\sqrt{5}}, \cos y = \frac{2}{\sqrt{5}} \)
from (i), \( \Rightarrow x \times \left( \frac{8}{5\sqrt{5}} + 3 \cdot \frac{2}{\sqrt{5}} \times \frac{1}{5} \right) = 14 \)
\( \Rightarrow x = 5\sqrt{5} \)
Case - II: y in IIIrd quadrant
\( \therefore \sin y = -\frac{1}{\sqrt{5}}, \cos y = -\frac{2}{\sqrt{5}} \)
from (i), \( \Rightarrow x = -5\sqrt{5} \)

Question. Solve the equation : \( \cot x - 2 \sin 2x = 1 \).
Answer: \( \cot x - 2 \sin 2x = 1 \quad \sin x \neq 0 \)
\( \frac{\cos x}{\sin x} - 2 \cdot 2 \sin x \cos x = 1 \)
\( \Rightarrow \cos x - 4\sin^2 x \cos x = \sin x \)
\( \Rightarrow \cos x - 4(1 - \cos^2 x) \cos x = \sin x \)
\( \Rightarrow \cos x - 4\cos x + 4 \cos^3 x = \sin x \)
\( \Rightarrow 4 \cos^3 x - 3\cos x = \sin x \)
\( \Rightarrow \cos 3x = \sin x \Rightarrow \cos 3x = \cos \left(\frac{\pi}{2} - x\right) \)
\( \Rightarrow 3x = 2n\pi \pm \left(\frac{\pi}{2} - x\right) \)
\( \Rightarrow 3x = 2n\pi + \frac{\pi}{2} - x \text{ or } 3x = 2n\pi - \frac{\pi}{2} + x \)
\( \Rightarrow 4x = 2n\pi + \frac{\pi}{2} \text{ or } 2x = 2n\pi - \frac{\pi}{2} \)
\( \Rightarrow x = \frac{n\pi}{2} + \frac{\pi}{8}, n \in I \text{ or } x = n\pi - \frac{\pi}{4}, n \in I \)
or \( x = n\pi + \frac{3\pi}{4}, n \in I \)

Question. Find real values of \( x \) for which, \( 27^{\cos 2x} \cdot 81^{\sin 2x} \) is minimum. Also find this minimum value.
Answer: Let \( y = 27^{\cos 2x} \cdot 81^{\sin 2x} = 3^{3\cos 2x + 4\sin 2x} \). The minimum value of the exponent \( 3\cos 2x + 4\sin 2x \) is \( -\sqrt{3^2 + 4^2} = -5 \). Thus, \( y_{\text{min}} = 3^{-5} = \frac{1}{243} \). This occurs when \( 3\cos 2x + 4\sin 2x = -5 \), which gives \( \cos(2x - \alpha) = -1 \) where \( \tan \alpha = 4/3 \). Solving this, \( 2x - \alpha = (2n+1)\pi \), so \( x = (2n + 1)\frac{\pi}{2} + \frac{\alpha}{2} \), where \( n \in \mathbb{I} \) and \( \alpha = \tan^{-1}(4/3) \).

Question. Solve the following system of equation for \( x \) and \( y \):
\( 5^{(\text{cosec}^2 x - 3\text{sec}^2 y)} = 1 \) and \( 2^{(2\text{cosec} x + \sqrt{3}|\text{sec} y|)} = 64 \).
Answer: From the first equation, \( \text{cosec}^2 x - 3\text{sec}^2 y = 0 \Rightarrow \text{cosec}^2 x = 3\text{sec}^2 y \Rightarrow \sqrt{3}|\text{sec} y| = |\text{cosec} x| \). Substituting into the second equation \( 2\text{cosec} x + \sqrt{3}|\text{sec} y| = 6 \), we find \( \text{cosec} x = 2 \) (since \( \text{cosec} x = 6/3 \)), so \( \sin x = 1/2 \Rightarrow x = n\pi + (-1)^n \frac{\pi}{6} \). For \( y \), \( 3\text{sec}^2 y = 4 \Rightarrow \cos^2 y = 3/4 \Rightarrow y = n\pi \pm \frac{\pi}{6} \), where \( n \in \mathbb{I} \).

Question. The number of integral values of \( k \) for which the equation \( 7 \cos x + 5 \sin x = 2k + 1 \) has a solution is
(a) 4
(b) 8
(c) 10
(d) 12
Answer: (b) 8

Question. For \( 0 < \theta < \pi/2 \), then solution(s) of \( \sum_{m=1}^{6} \text{cosec} (\theta + (m - 1)\pi/4) \text{cosec}(\theta + m\pi/4) = 4\sqrt{2} \) is(are)
(a) \( \pi/4 \)
(b) \( \pi/6 \)
(c) \( \pi/12 \)
(d) \( 5\pi/12 \)
Answer: (c) \( \pi/12 \) and (d) \( 5\pi/12 \)

Question. The number of values of \( \theta \) in the interval \( (-\pi/2, \pi/2) \) such that \( \theta \neq \frac{n\pi}{5} \) for \( n=0, \pm 1, \pm 2 \) and \( \tan \theta = \cot 5\theta \) as well as \( \sin 2\theta = \cos 4\theta \) is
Answer: 3

Question. Let \( P = \{ \theta : \sin \theta - \cos \theta = \sqrt{2} \cos \theta \} \) and \( Q = \{ \theta : \sin \theta + \cos \theta = \sqrt{2} \sin \theta \} \) be two sets. Then
(a) \( P \subset Q \) and \( Q - P \neq \emptyset \)
(b) \( Q \not\subset P \)
(c) \( P \not\subset Q \)
(d) \( P = Q \)
Answer: (d) \( P = Q \)