Question

The number of solution(s) of the equation $2{\cos }^{2}x-2\sqrt{2}\cos x+{\tan }^{2}x-2\tan x+2=0$ where $x\in [-4\pi ,4\pi ]$ is

Answer 1

$2{\cos }^{2}x-2\sqrt{2}\cos x+{\tan }^{2}x-2\tan x+2=0$
$\Rightarrow (\sqrt{2}\cos x{)}^2-2\sqrt{2}\cos x+1+{\tan }^{2}x-2\tan x+1=0$
$\Rightarrow (\sqrt{2}\cos x-1{)}^2+(\tan x-1{)}^2=0$
This is only possible when
$\sqrt{2}\cos x-1=0\text{and}\tan x-1=0$
$\Rightarrow \cos x=\frac{1}{\sqrt{2}}\text{and}\tan x=1\Rightarrow x=2n\pi \pm \frac{\pi }{4}\text{and}x=n\pi +\frac{\pi }{4},n\in Z$
So, in the given interval solutions are
$-\frac{15\pi }{4},-\frac{7\pi }{4},\frac{\pi }{4},\frac{9\pi }{4}$

Hence, the number of solution is $4$.