Solve the following equations:

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

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

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

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

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

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

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

$\Rightarrow \cos x=\frac{1}{2}$ or $\cos x=-\frac{3}{2}$

$\cos x=-\frac{3}{2}$ is not possible.

$\therefore \cos x=\frac{1}{2}$

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

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

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

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

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

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

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

$\cos x=2$ is not possible.

$\therefore 2 \cos x-1=0$

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

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

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

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

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

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

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

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

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

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

$\cos x=\sqrt{3}$ is not possible.

$\therefore 2 \cos x+\sqrt{3}=0$

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

$\Rightarrow \cos x=\cos \frac{5 \pi}{6}$

$\Rightarrow x=2 n \pi \pm \frac{5 \pi}{6}, \mathrm{n} \in$

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

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

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

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

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

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

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

Now,

$2 \cos x+5=0 \Rightarrow \cos x=-\frac{5}{2}$ (It is not possible.)

$\therefore 2 \cos x-1=0$

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

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

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

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

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

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

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

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

Now,

$\tan x-\sqrt{3}=0$

$\Rightarrow \tan x=\sqrt{3}$

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

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

And,

$\tan x=-1$

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

$\Rightarrow x=m \pi-\frac{\pi}{4}, \mathrm{~m} \in \mathrm{Z}$

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

Now,

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

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

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

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

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

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

$\Rightarrow 2 x=n \pi+\frac{\pi}{3}, n \in Z$

$\Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{6}, n \in Z$

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

$\Rightarrow 4 x=2 n \pi \pm 2 x, n \in Z$

On taking positive sign, we have:

$4 x=2 n \pi+2 x$

$\Rightarrow 2 x=2 n \pi$

$\Rightarrow x=n \pi, n \in Z$

On taking negative sign, we have:

$4 x=2 n \pi-2 x$

$\Rightarrow 6 x=2 n \pi$

$\Rightarrow x=\frac{n \pi}{3}, n \in Z$