$\cos 2x-2\tan x+2=0$

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

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

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

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

Putting $\tan x=t$, we get

$2t^3-t^2+2t-3=0$

$\Rightarrow \left(t-1\right)\left(2t^2+t+3\right)=0$

Hence, $t=1$ is the only solution

$\Rightarrow \tan x=1=\tan \frac{\pi }{4}$

$\Rightarrow \theta =n\pi +\frac{\pi }{4}, n\in I$.