24. Trigonometrical Equations and their General Solutions Answers#

Given equation is $\sin\theta = -1$

$\Rightarrow \sin\theta = \sin\left(-\frac{\pi}{2}\right)$

$\Rightarrow \theta = n\pi + (-1)^n\left(-\frac{\pi}{2}\right)$

$\theta = n\pi + (-1)^{n +1}\frac{\pi}{2}$ where $n\in I.$
Given equation is $\cos\theta = -\frac{1}{2}$

$\cos\theta = \cos\frac{2\pi}{3} \Rightarrow \theta = 2n\pi \pm \frac{2\pi}{3}$ where $n\in I.$
Given equation is $\tan\theta = -\sqrt{3}$

$\Rightarrow \tan\theta = \tan\left(-\frac{\pi}{3}\right) \Rightarrow \theta = n\pi + \left(-\frac{\pi}{3}\right)$

$= n\pi - \frac{\pi}{3}$ where $n\in I.$
Given equation is $\sec\theta = -\sqrt{2}$

$\Rightarrow \sec\theta = \sec\frac{3\pi}{4}\Rightarrow \theta = 2n\pi\pm \frac{3\pi}{4}$ where $x\in I$
Given equation is $\sin8\theta = \sin\theta \Rightarrow \sin9\theta - \sin\theta = 0$

$\Rightarrow 2\cos5\theta.\sin4\theta = 0$

Either $\cos5\theta = 0$ or $\sin4\theta = 0$

$\Rightarrow 5\theta = (2n + 1)\frac{\pi}{2}$ or $4\theta = n\pi$

$\theta = \frac{n\pi}{4}, (2n + 1)\frac{\pi}{10}$ where $n\in I.$
Given equation is $\sin5x = \cos2x$

$\Rightarrow \cos2x = \cos\left(\frac{\pi}{2} - 5x\right)$

$2x = 2n\pi \pm \left(\frac{\pi}{2} - 5x\right)$

$x = (4n + 1)\frac{\pi}{14}, -(4n - 1)\frac{\pi}{6}$ where $n\in I$
Given equation is $\sin3x = \sin x \Rightarrow \sin3x - \sin x= 0$

$\Rightarrow \cos2x.\sin x = 0$

Either $\cos2x = 0$ or $\sin x= 0$

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

$x = n\pi, (2n + 1)\frac{\pi}{4}$ where $n\in I$
Given equation is $\sin3x = \cos2x \Rightarrow \cos2x = \cos\left(\frac{\pi}{2} - 3x\right)$

$\Rightarrow 2x = 2n\pi \pm \left(\frac{\pi}{2} - 3x\right)$

$x = \frac{2n\pi}{5} + \frac{\pi}{10}, -2n\pi + \frac{\pi}{2}$
Given equation is $\sin ax + \cos bx = 0$

$\Rightarrow \sin ax + \sin\left(\frac{\pi}{2} - bx\right) = 0$

$\Rightarrow 2\sin\left(\frac{\pi}{4} + \frac{(a - b)x}{2}\right)\cos\left(\frac{(a + b)x}{2} - \frac{\pi}{4}\right) = 0$

Either $\Rightarrow \sin\left(\frac{\pi}{4} + \frac{(a - b)x}{2}\right) = 0$ or $\cos\left(\frac{(a + b)x}{2} - \frac{\pi}{4}\right) = 0$

$\Rightarrow \frac{\pi}{4} + \frac{(a - b)x}{2} = n\pi$ or $\frac{(a + b)x}{2} - \frac{\pi}{4} = (2n + 1)\frac{\pi}{2}$

$x = \frac{2n\pi - \frac{\pi}{2}}{a - b}, \frac{(2n + 1)\pi + \frac{\pi}{2}}{a + b}$
Given $\tan x\tan 4x =1 \Rightarrow \sin x\sin4x = \cos x\cos4x$

$\Rightarrow \cos x\cos4x - \sin x\sin 4x = 0$

$\cos 5x = 0 \Rightarrow 5x = (2n + 1)\frac{\pi}{2} \Rightarrow x = \frac{(2n + 1)\pi}{10}$
Given equation is $\cos\theta = \sin105^\circ + \cos 105^\circ$

$\sin105^\circ = \sin(60^\circ + 45^\circ) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

$\cos105^\circ = \cos(60^\circ + 45^\circ) = \frac{1 - \sqrt{3}}{2\sqrt{2}}$

$\Rightarrow \cos\theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = 2n\pi \pm \frac{\pi}{4}$
Given equation is $7\cos^2\theta + 3\sin^2\theta = 4$

$\Rightarrow 4\cos^2\theta + 3 = 4 \Rightarrow \cos\theta = \pm \frac{1}{2}$

If $\cos\theta = \frac{1}{2}\Rightarrow \theta = 2n\pi\pm\frac{\pi}{3}$

If $\cos\theta = -\frac{1}{2}\Rightarrow \theta = 2n\pi\pm\frac{2\pi}{3}$
Given equation is $3\tan(\theta - 15^\circ) = \tan(\theta + 15^\circ)$

$\Rightarrow \frac{\tan(\theta + 15^\circ)}{\tan(\theta - 15^\circ)} = \frac{3}{1}$

Applying componendo and dividendo

$\Rightarrow \frac{\tan(\theta + 15^\circ) + \tan(\theta - 15^\circ)}{\tan(\theta + 15^\circ) - \tan(\theta - 15^\circ)} = \frac{4}{2}$

$\Rightarrow \frac{\sin(\theta + 15^\circ + \theta - 15^\circ)}{\sin(\theta + 15^\circ - \theta + 15^\circ)} = 2$

$\Rightarrow \sin2\theta = 2 \Rightarrow 2\theta = n\pi + (-1)^n\frac{\pi}{2} \Rightarrow \theta = \frac{n\pi}{2} + (-1)^n\frac{\pi}{4}$
Given equation is $\tan x + \cot x = 2 \Rightarrow \tan^2x - 2\tan x + 1 = 0$

$\Rightarrow (\tan x - 1)^2 = 0 \Rightarrow \tan x = 1 \Rightarrow x = n\pi + \frac{\pi}{4}$
Given equation is $\sin^2\theta = \sin^2\alpha \Rightarrow \sin\theta = \pm \sin\alpha$

$\theta = n\pi \pm\alpha$
Given equation is $\tan^2x + \cot^2x = 2$

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

$\tan x = \pm \Rightarrow x = n\pi \pm \frac{\pi}{4}$
Given equation is $\tan^2x = 3\cosec^2x - 1$

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

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

If $\tan^2x + 1 = 0$ then $x$ will become imaginary.

$\therefore \tan x = \pm\sqrt{3} \Rightarrow x = n\pi \pm \frac{\pi}{3}$
Given equation is $2\sin^2x + \sin^22x = 2$

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

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

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

$\Rightarrow x = n\pi \pm \frac{\pi}{4}, (2n + 1)\frac{\pi}{2}$
Given equation is $7\cos^2 x + 3\sin^2 x = 4$

$\Rightarrow 4\cos^2 x + 3 = 4 \Rightarrow \cos x = \pm \frac{1}{2}$

If $\cos x = \frac{1}{2}\Rightarrow x = 2n\pi\pm\frac{\pi}{3}$

If $\cos x = -\frac{1}{2}\Rightarrow x = 2n\pi\pm\frac{2\pi}{3}$
Given equation is $2\cos2x + \sqrt{2\sin x} = 2$

$\Rightarrow \sqrt{2\sin x} = 2(1 - \cos2x) = 4\sin^2x$

$\Rightarrow \sqrt{2\sin x}\left(1 - 2\sqrt{2}\sin^{\frac{3}{2}}x\right) = 0$

Either:math:sin x = 0 Rightarrow x = npi where $n\in I$

or $\sin^{\frac{3}{2}}x = \frac{1}{2\sqrt{2}} \Rightarrow \sin x = \frac{1}{2}$

$\Rightarrow x = n\pi + (-1)^n\frac{\pi}{6}$
We know that $\tan^2\frac{x}{2} = \frac{1 - \cos x}{1 + \cos x}$

$\therefore 8\left(\frac{1 - \cos x}{1 + \cos x}\right) = 1 + sec x = \frac{1 + \cos x}{\cos x}$

$\Rightarrow 8\cos x - 8\cos^2x = (1 + \cos x)^2$

$\Rightarrow 9\cos^2x - 6\cos x + 1 = 0 \Rightarrow (3\cos x - 1)^2 = 0$

$\cos x = \frac{1}{3} \Rightarrow x = 2n\pi \pm \cos^{-1}\frac{1}{3}$ where $n in I.$

Check $\frac{x}{2}\neq (2n + 1)\frac{\pi}{2}$ and $\cos x \neq = 0$ else equation will be meaningless.

$\Rightarrow x\neq (2n + 1)\pi$ and $x\neq (2n + 1)\frac{\pi}{2}$
Given equation is $\cos x\cos2x\cos3x = \frac{1}{4}$

$\Rightarrow (2\cos x\cos3x)2\cos2x = 1 \Rightarrow 2\cos4x\cos2x + 2\cos^22x - 1 = 0$

$\Rightarrow \cos4x[2\cos2x + 1] = 0$

If $\cos4x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{8}$

If $2\cos2x + 1 = 0 \Rightarrow 2x = 2n\pi \pm \frac{2\pi}{3}$

$x = n\pi \pm \frac{\pi}{3}$
Given equation is $\tan x + \tan2x + \tan3x = 0$

$\Rightarrow \tan x +\tan2x + \frac{\tan x + \tan 2x}{1 - \tan x\tan 2x} = 0$

$\Rightarrow (\tan x + \tan 2x)\left(1 + \frac{1}{1 - \tan x\tan 2x}\right) = 0$

If $\tan x + \tan 2x = 0 \Rightarrow \tan x = -\tan 2x \Rightarrow x = n\pi -2x \Rightarrow x = \frac{n\pi}{3}$

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

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

$x = n\pi \pm\tan^{-1}\frac{1}{\sqrt{2}}$
Given equation is $\cot x - \tan x - \cos x + \sin x = 0$

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

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

If $\cos x - \sin x = 0 \Rightarrow \tan x = 1\Rightarrow x = n\pi + \frac{\pi}{4}$

If $\frac{\cos x + \sin x}{\cos x\sin x} - 1= 0$

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

Squaring, we get $1 + \sin2x = \frac{1}{4}\sin^2x$

$\Rightarrow \sin2x = 2\pm 2\sqrt{2}$

However, $2 + 2\sqrt{2} > 1$ which is not possible.

$\Rightarrow \sin 2x = 2 - 2\sqrt{2} = \sin\alpha$ (let)

$x = \frac{n\pi}{2} + \frac{(-1)^n\alpha}{2}$
Given equation is $2\sin^2x - 5\sin x\cos x - 8\cos^2x = -2$

Clearly, $\cos x \neq 0$ else $\sin^2x = -1$ which is not possible.

Therefore, we can divide both sides by $\cos^2x$ which yields

$2\tan^2x - 5\tan x -8 = -2\sec^2x$

$\Rightarrow 4\tan^2x - 5\tan x - 6 = 0$

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

Thus, $x = n\pi + \tan^{-1}2, b\pi + \tan^{-1}\left(\frac{-3}{4}\right)$
Given equation is $(1 - \tan x)(1 + \sin2x) = 1 + \tan x$

$\Rightarrow (1 - \tan x)\left(1 + \frac{2\tan x}{1 + \tan^2x}\right) = 1 + \tan x$

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

$\Rightarrow (1 + \tan x)[(1 - \tan x)(1 + \tan x) - (1 + \tan^2x)] = 0$

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

If $\tan^2x = 0 \Rightarrow \tan x = 0\Rightarrow x = n\pi$

If $1 + \tan x = 0 \Rightarrow x = n\pi - \frac{\pi}{4}$

where $n \in I$
Given equation is $2(\cos x + \cos2x) + \sin2x(1 + 2\cos x) = 2\sin x$

$\Rightarrow 4\cos\frac{3x}{2}\cos\frac{x}{2} + 2\sin\frac{5x}{2}\cos\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = 0$

$\Rightarrow 2\cos\frac{x}{2}\left[2\cos\frac{3x}{2} + \sin\frac{5x}{2} - \sin\frac{x}{2}\right] = 0$

$\Rightarrow 2\cos\frac{x}{2}\left[2\cos\frac{3x}{2} + 2\cos\frac{3x}{2}\sin x\right] = 0$

$\Rightarrow 4\cos\frac{x}{2}\cos\frac{3x}{2}[1 + \sin x] = 0$

If $\cos\frac{x}{2} = 0 \Rightarrow x = (2n + 1)\pi$

If $\cos\frac{3x}{2} = 0 \Rightarrow x = (2n + 1)\frac{\pi}{3}$

If $1 + \sin x = 0 \Rightarrow x = n\pi + (-1)^{n + 1}\frac{\pi}{2}$

So the values in the given range are $x = -\pi, -\frac{\pi}{2}, -\frac{\pi}{3}, \frac{\pi}{3}, \pi$
Given equation is $4\cos^2x\sin x - 2\sin^2x = 3\sin x$

$\Rightarrow \sin x[4\cos^2x - 2\sin x - 3] = 0$

$\Rightarrow \sin x[4 - 4\sin^2x - 2\sin x - 3] = 0$

$\Rightarrow \sin x[4\sin^2x + 2\sin x - 1] = 0$

If $\sin x = 0 \Rightarrow x = n\pi$

If $4\sin^2x + 2\sin x - 1 = 0$

$\sin x = \frac{-1\pm\sqrt{5}}{4}$

If $\sin x = \frac{-1 + \sqrt{5}}{4} \Rightarrow \sin x = \sin\frac{\pi}{10} \Rightarrow x = n\pi + (-1)^n\frac{\pi}{10}$

If $\sin x = \frac{-1 - \sqrt{5}}{4}\Rightarrow \sin x = -\sin54^\circ = \sin\left(\frac{-3\pi}{10}\right)$

$\Rightarrow x = n\pi + (-1)^{n + 1}\frac{3\pi}{10}$
Given equation is $2 + 7\tan^2x = 3.25\sec^2x$

$\Rightarrow 8 + 28\tan^2x = 13\sec^2x = 13 + 13\tan^2x$

$\Rightarrow 15\tan^2x = 5 \Rightarrow \tan x = \pm \frac{1}{\sqrt{3}}$

$\Rightarrow x = n\pi \pm \frac{\pi}{6}$
Given equation is $\cos 2x + \cos 4x = 2\cos x$

$\Rightarrow \cos4x + \cos2x - 2\cos x = 0$

$\Rightarrow 2\cos3x\cos x - \cos x = 0$

$2\cos x[\cos 3x - 1] = 0$

If $\cos x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{2}$

If $\cos 3x - 1 = 0\Rightarrow 3x = 2n\pi \Rightarrow x = \frac{2n\pi}{3}$
Given equation is $3\tan x + \cot x = 5\cosec x$

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

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

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

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

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

$\sin x\neq 0$ because that will make $\cosec x$ and $\cot x \infty.$

$2\cos x + 3 \neq 0$ because $-1\leq \cos x\leq 1$

$\therefore 2\cos x - 1 = 0\Rightarrow \cos x = \frac{1}{2} \Rightarrow x = 2n\pi \pm \frac{\pi}{3}$
Given equation is $2\sin^2x = 3\cos x$

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

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

$\cos x \neq 2 \because -1\leq \cos x\leq 1$

$\therefore 2\cos x - 1 = 0\Rightarrow x = 2n\pi \pm \frac{\pi}{3}~\forall~n\in I$

$0\leq x \leq 2\pi \therefore x = \frac{\pi}{3}, \frac{5\pi}{3}$
Given equation is $\sin^2x - \cos x = \frac{1}{4}$

$\Rightarrow 4\sin^2x - 4\cos x = 1 \Rightarrow 4 - 4\cos^2x - 4\cos x = 1$

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

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

$\cos x \neq 2 \because -1\leq \cos x\leq 1$

$\therefore 2\cos x - 1 = 0\Rightarrow x = 2n\pi \pm \frac{\pi}{3}~\forall~n\in I$

$0\leq x \leq 2\pi \therefore x = \frac{\pi}{3}, \frac{5\pi}{3}$
Given equation is $3\tan^2x - 2\sin x = 0$

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

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

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

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

$\sin x\neq -2 \because -1\leq \sin x\leq 1$

If $\sin x = 0 \Rightarrow x = n\pi$

If $2\sin x - 1 = 0 \Rightarrow x = n\pi + (-1)^n\frac{\pi}{6}$
Given equation is $\sin x + \sin5x = \sin 3x$

$\Rightarrow \sin5x - \sin3x + \sin x = 0$

$\Rightarrow 2\cos4x\sin x + \sin x = 0$

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

If $\sin x = 0\Rightarrow x = n\pi~\forall~x\in I$

If $2\cos4x + 1 = 0 \Rightarrow 4x = 2n\pi \pm\frac{2\pi}{3}$

$x = \frac{n\pi}{2}\pm \frac{\pi}{6}~\forall~x\in I$

Thus, $x = 0, \pi$ and $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}$
Given equation is $\sin6x = \sin4x - \sin2x$

$\Rightarrow \sin6x + \sin2x - \sin4x = 0$

$\Rightarrow 2\sin4x\cos2x - \sin4x = 0$

$\Rightarrow \sin4x(2\cos2x - 1) = 0$

If $\sin4x = 0 \Rightarrow x = \frac{n\pi}{4}$

If $2\cos2x - 1 = 0 \Rightarrow \cos2x = \frac{1}{2}\Rightarrow 2x = 2n\pi \pm \frac{\pi}{3}$

$\Rightarrow x = n\pi \pm \frac{\pi}{6}$
Given equation is $\cos6x + \cos 4x + \cos 2x + 1 = 0$

$\Rightarrow 2\cos5x\cos x + 2\cos^2x = 0$

$\Rightarrow 2\cos x(\cos5x + \cos x) = 0$

$\Rightarrow 4\cos x\cos2x\cos3x = 0$

If $\cos x = 0 \Rightarrow x = 2n\pi \pm\frac{\pi}{2}$

If $\cos2x = 0 \Rightarrow x = n\pi \pm \frac{\pi}{4}$

If $\cos3x = 0 \Rightarrow x = \frac{2n\pi}{3}\pm\frac{\pi}{6}$
Given equation is $\cos x + \cos 2x + \cos 3x = 0$

$\Rightarrow (\cos x + \cos3x) + \cos 2x = 0$

$\Rightarrow 2\cos2x\cos x + \cos 2x = 0$

$\Rightarrow \cos2x(2\cos x + 1) = 0$

If $\cos 2x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{4}$

If $2\cos x + 1 = 0 \Rightarrow x = 2n\pi\pm\frac{2\pi}{3}$
Given equation is $\cos3x + \cos2x = \sin\frac{3x}{2} + \sin\frac{x}{2}$

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

$\Rightarrow 2\cos\frac{x}{2}\left[\cos\frac{5x}{2} - \sin x\right] = 0$

If $\cos\frac{x}{2} = 0 \Rightarrow \frac{x}{2} = \left(n + \frac{1}{2}\right)\pi$

$x = (2n + 1)\pi$

If $\cos\frac{5x}{2} = \sin x = \cos\left(\frac{\pi}{2} - x\right)$

$\Rightarrow \frac{5x}{2} = 2n\pi \pm \left(\frac{\pi}{2} - x\right)$

$\Rightarrow x = (4n + 1)\pi/7, (4n - 1)\pi/3$

Thus, between $0$ and $2\pi$ the values of $x$ are $\frac{\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{13\pi}{7}.$
Given equation is $\tan x+ \tan2x + \tan3x = \tan x.\tan2x.\tan3x$

$\Rightarrow \tan x + \tan2x = \tan3x(\tan x\tan2x - 1)$

$\Rightarrow \frac{\tan x + \tan 2x}{1 - \tan x\tan2x} = -\tan3x$

$\Rightarrow \tan(x + 2x) = -\tan3x \Rightarrow 2\tan 3x = 0$

$\Rightarrow 3x = n\pi \Rightarrow x = \frac{n\pi}{3}$
Given equation is $\tan x + \tan 2x + \tan x\tan 2x = 1$

$\Rightarrow \tan x + \tan2x = 1 - \tan x\tan 2x$

$\Rightarrow \frac{\tan x + \tan2x}{1 - \tan x\tan2x} = 1$

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

$3x = n\pi + \frac{\pi}{4} \Rightarrow x = (4n + 1)\frac{\pi}{12}$
Given equation is $\sin2x + \cos2x + \sin x + \cos x + 1 = 0$

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

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

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

If $\cos x = -\frac{1}{2} \Rightarrow x = 2n\pi\pm\frac{2\pi}{3}$

If $\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x = n\pi - \frac{\pi}{4}$
We have to prove that $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$

$\Rightarrow (\sin x + \sin 3x) + \sin 2x = (\cos x + \cos3x) + \cos 2x$

$\Rightarrow 2\sin2x\cos x + \sin 2x = 2\cos2x\cos x + \cos 2x$

$\Rightarrow \sin2x(2\cos x + 1) = \cos2x(2\cos x + 1)$

$\Rightarrow (2\cos x + 1)(\sin2x - \cos2x) = 0$

If $2\cos x + 1 = 0 \Rightarrow x = 2n\pi \pm \frac{2\pi}{3}$

If $\sin 2x - \cos2x = 0 \Rightarrow \tan2x = 1 = \tan\frac{\pi}{4} \Rightarrow x = \frac{n\pi}{2}+\frac{\pi}{8}$
Given equation is $\cos6x + \cos4x = \sin3x + \sin x$

$\Rightarrow 2\cos5x\cos x = 2\sin2x\cos x$

$\Rightarrow \cos x(\cos5x - \sin2x) = 0$

If $x = 0 \Rightarrow x = 2n\pi\pm\frac{\pi}{2}$

If $\cos5x = \sin2x \Rightarrow \cos5x = \cos\left(\frac{\pi}{2} - 2x\right)$

$\Rightarrow 5x = 2n\pi\pm \left(\frac{\pi}{2} - 2x\right)$

Taking +ve sign $7x = 2n\pi + \frac{\pi}{2} \Rightarrow x = (4n + 1)\frac{\pi}{14}$

Taling -ve sign $3x = 2n\pi - \frac{\pi}{2} \Rightarrow x = (4n - 1)\frac{\pi}{6}$
Given equation is $\sec4x - \sec2x = 2$

$\Rightarrow \cos2x - \cos4x = 2\cos2x\cos4x$ where $\cos2x, \cos4x\neq 0$

$\Rightarrow \cos2x - \cos4x = \cos6x + \cos 2x$

$\Rightarrow \cos6x + \cos4x = 0 \Rightarrow 2\cos5x\cos x = 0$

If $\cos 5x = 0 \Rightarrow 5x = 2n\pi\pm\frac{\pi}{2}\Rightarrow x = \frac{2n\pi}{5}\pm\frac{\pi}{10}$

If $\cos x = 0 \Rightarrow x = 2n\pi\pm\frac{\pi}{2}$
Given equation is $\cos2x = (\sqrt{2} + 1)\left(\cos x - \frac{1}{\sqrt{2}}\right)$

$\Rightarrow (2\cos^2x - 1) = \frac{\sqrt{2} + 1}{\sqrt{2}}\left(\sqrt{2}\cos x - 1\right)$

$\Rightarrow (\sqrt{2}\cos x - 1)\left(\sqrt{2}\cos x + 1 - 1 - \frac{1}{\sqrt{2}}\right) = 0$

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

If $\sqrt{2}\cos x - 1 = 0 \Rightarrow x = 2n\pi \pm \frac{\pi}{4}$

If $2\cos x - 1 = 0 \Rightarrow x = 2n\pi\pm\frac{\pi}{3}$
Given equation is $5\cos2x + 2\cos^2\frac{x}{2} + 1 = 0$

$\Rightarrow 10\cos^2x - 5 + \cos x + 2 = 0[\because \cos2x = 2\cos^2x - 1]$

$\Rightarrow 10\cos^2x + \cos x - 3 = 0$

$\Rightarrow (2\cos x -1)(5\cos x + 3) = 0$

If $\cos x = 1/2 \Rightarrow x = \frac{\pi}{3} [\because -\pi \leq x\leq \pi]$

If $5\cos x + 3 = 0 \Rightarrow x = \pi - \cos^{-1}\frac{3}{5}$
Given equation is $\cot x - \tan x = sec x$

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

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

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

$\cos x\neq 0$ and $\sin \neq -1$ because that will render original equation meaningless.

$\therefore 2\sin x - 1= 0 \Rightarrow x = n\pi + (-1)^n\frac{\pi}{6}$
Given equation is $1 + \sec x = \cot^2\frac{x}{2}$

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

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

$\Rightarrow \cos^2\frac{x}{2}\left(2\sin^2\frac{x}{2} - \cos x\right) = 0$

$\Rightarrow \cos^2\frac{x}{2}\left(1 - 2\cos x\right) = 0$

If $\cos\frac{x}{2} = 0 \Rightarrow \frac{x}{2} = n\pi + \frac{pi}{2} \Rightarrow x = (2n + 1)\pi$

If $1 - 2\cos x = 0\Rightarrow x = 2n\pi \pm \frac{\pi}{3}$
Given equation is $\cos3x\cos^3x + \sin3x\sin^3x = 0$

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

$\Rightarrow 3(\sin^4x - \cos^4x) - 4(\sin^6x - \cos^6x) = 0$

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

$\Rightarrow \cos2x[-3 + 4\{\sin^2x(\sin^2x + \cos^2x) + \cos^4x\}] = 0$

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

$\Rightarrow \cos2x(2\cos^2x - 1)^2 = 0$

$\Rightarrow \cos^32x = 0$

$\Rightarrow \cos 2x = 0$

$2x = n\pi + \frac{\pi}{2} \Rightarrow x = (2n + 1)\frac{\pi}{4}$
Given equation is $\sin^3x + \sin x\cos x + \cos^3x = 1$

$\Rightarrow \sin^3x + \cos^3x + \sin x\cos x - 1 = 0$

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

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

If $1 - \sin x\cos x = 0 \Rightarrow \sin 2x = 2$ which is not possible.

$\therefore \sin x + \cos x - 1 = 0 \Rightarrow \frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{\sqrt{2}}$

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

$\Rightarrow x - \pi/4 = 2n\pi \pm \pi/4 = 2n\pi, 2n\pi + \pi/2$
Given equation is $\sin 7x + \sin4x + \sin x = 0$

$\Rightarrow 2\sin4x\cos3x + \sin4x = 0$

$\Rightarrow \sin4x(2\cos3x + 1) = 0$

If $\sin4x = 0 \Rightarrow x = n\pi/4 \Rightarrow x = \pi/4~\forall 0\leq x\leq\pi/2$

If $\cos3x = -1/2 \Rightarrow x = \frac{2\pi}{9}, \frac{4\pi}{9}~\forall 0\leq x\leq\pi/2$
Given equation is $\sin x + \sqrt{3}\cos x = \sqrt{2}$

Dividing both sides by $2$ [we arrive at this no. by squaring and adding coefficients of $\sin x$ and $\cos x$ and then taking square root]

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

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

$\Rightarrow \cos\left(x - \frac{\pi}{6}\right) = \cos\frac{\pi}{4}$

$\Rightarrow x - \frac{\pi}{6} = 2n\pi\pm\frac{\pi}{4}$

$\Rightarrow x = 2n\pi\pm\frac{5\pi}{12}, 2n\pi-\frac{\pi}{12}$
We have to find minimum value of $27^{\cos2x}.81^{\sin2x}$