Question

2-sin^2x=2cos^2(x/2)

Answer 1

$2-\sin^2x=2\cos^2\dfrac x2$
$1+\cos^2x=1+\cos x$
$\cos^2x-\cos x=0$
$\cos x(\cos x-1)=0$

and this has solutions of $x=\dfrac{n\pi}2=\dfrac{(2k+1)\pi}2$ and $x=n\pi=2k\pi$ where $k\in\mathbb Z$.