Корень их двух cos квадрат x=sin(x-П/2

Решите задачу:

$\sqrt2cos^2x=sin(x-\frac{\pi}{2})\\\\\sqrt2cos^2x=-sin(\frac{\pi}{2}-x)\\\\\sqrt2cos^2x=-cosx\\\\\sqrt2cos^2x+cosx=0\\\\cosx(\sqrt2cosx+1)=0\\\\a)\; \; cosx=0,\; \; x=\frac{\pi}{2}+\pi n,\; n\in Z\\\\b)\; \; cosx=-\frac{1}{\sqrt2}\\\\x=\pm arccos(-\frac{1}{\sqrt2})+2\pi n=\pm (\pi -\frac{\pi}{4})+2\pi n=\pm \frac{3\pi}{4}+2\pi n$