${ cos }^{ 2 }2x+2{ cos }^{ 2 }x=1,x\epsilon \left( -\pi ,\pi ; \right)$, then $x$ can take the values

Given $x\in (-\pi,\pi)\implies 2 x\in (-2\pi,2\pi)$

$\cos^2 2 x+2\cos^2 x=1$

$\implies \cos^2 2 x+2\cos^2 x-1=0$

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

$\implies \cos^2 2 x+\cos 2 x=0$

$\implies \cos 2 x(\cos 2 x+1)=0$

So $\cos 2 x=0$ or $-1$

$\implies 2 x=\pm \dfrac{\pi}{2},\pm \dfrac{3\pi}{2}$ or $\pm \pi$

$\implies x=\pm \dfrac{\pi}{4},\pm \dfrac{3\pi}{4}$ or $\pm\dfrac{\pi}{2}$