Question

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

Answer 1

`2-\sin^2x=2\cos^2\frac x2`

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

`\cos^2x-\cos x=0`

`\cos x(\cos x-1)=0`

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