Вопрос в картинках...

Решите задачу:

$\sqrt2\sin^2\left(\dfrac{\pi}{2}+x\right)=\sin\left(\dfrac{3\pi}{2}+x\right)\\ \sqrt2 \cos^2 x=-\cos x\\\ \sqrt2\cos^2 x+\cos x=0\\ \cos x(\sqrt 2\cos x+1)=0\\\\ \cos x=0\\ x=\dfrac{\pi}{2}+k\pi\\\\ \sqrt 2\cos x+1=0\\ \sqrt 2 \cos x=-1\\ \cos x=-\dfrac{1}{\sqrt2}\\ \cos x=-\dfrac{\sqrt2}{2}\\ x=-\dfrac{3\pi}{4}+2k\pi \vee x=\dfrac{3\pi}{4}+2k\pi \\\\ \boxed{x=\left\{\dfrac{\pi}{2}+k\pi,-\dfrac{3\pi}{4}+2k\pi,x=\dfrac{3\pi}{4}+2k\pi,(k\in\mathbb{Z})\right\}}$