Question

Number of values of $x$ satisfying $2{\sin }^{2}x+{\sin }^{2}2x=2$ and $\sin 2x+\cos 2x=\tan x$, where $x\in [-\frac{3\pi }{2},\frac{4\pi }{3}]$ is

Answer 1

$2{\sin }^{2}x+{\sin }^{2}2x=2\Rightarrow {\sin }^{2}2x=2{\cos }^{2}x\Rightarrow 4{\sin }^{2}x{\cos }^{2}x=2{\cos }^{2}x\Rightarrow {\cos }^{2}x=0\text{or}2{\sin }^{2}x=1$
When ${\cos }^{2}x=0$
$\Rightarrow x=\{-\frac{3\pi }{2},-\frac{\pi }{2},\frac{\pi }{2}\}$
and when ${\sin }^{2}x=\frac{1}{2}$
$\Rightarrow x=\{\pm \frac{\pi }{4},\pm \frac{3\pi }{4},\pm \frac{5\pi }{4}\}$
$\therefore x=\{-\frac{3\pi }{2},-\frac{5\pi }{4},-\frac{3\pi }{4},-\frac{\pi }{2},-\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2},\frac{3\pi }{4},\frac{5\pi }{4}\}$
For
$\sin 2x+\cos 2x=\tan x\Rightarrow \frac{2\tan x+1-{\tan }^{2}x}{1+{\tan }^{2}x}=\tan x\Rightarrow {\tan }^{3}x+{\tan }^{2}x-\tan x-1=0\Rightarrow ({\tan }^{2}x-1)(\tan x+1)=0\Rightarrow \tan x=-1,1\Rightarrow x=\{\pm \frac{\pi }{4},\pm \frac{3\pi }{4},\pm \frac{5\pi }{4}\}$
$\therefore$ no of common solutions is $6$