[tex]2-\sin^2x=2\cos^2\dfrac x2[/tex]
[tex]1+\cos^2x=1+\cos x[/tex]
[tex]\cos^2x-\cos x=0[/tex]
[tex]\cos x(\cos x-1)=0[/tex]
and this has solutions of [tex]x=\dfrac{n\pi}2=\dfrac{(2k+1)\pi}2[/tex] and [tex]x=n\pi=2k\pi[/tex] where [tex]k\in\mathbb Z[/tex].
