Помогите найти корни уравнения.

$\sqrt{2}\sin^3 x-\sqrt{2}\sin x+\cos^2 x=0; \ \sin x=t\in[-1;1];$

$\sqrt{2}t^3-\sqrt{2}t+1-t^2=0; \ 2\sqrt{2}t^3-2t^2-2\sqrt{2}t+2=0; \sqrt{2}t=p;$

$p^3-p^2-2p+2=0; p^2(p-1)-2(p-1)=0; (p^2-2)(p-1)=0;$

$p=1; p=\pm\sqrt{2}; t=\frac{1}{\sqrt{2}}; t=\pm 1; \sin x=\frac{\sqrt{2}}{2}; \sin x =\pm 1;$

$x=\frac{\pi}{4}+2\pi n; x=\frac{3\pi}{4}+2\pi m; x=\frac{\pi}{2}+\pi k; n,m,k\in Z$

Отбор:
$-\frac{5\pi}{2} \leq \frac{\pi}{4}+2\pi n \leq -\pi;\ -10 \leq 1+8n \leq -4;\ -\frac{11}{8} \leq n \leq -\frac{5}{8}\Rightarrow n=-1;$

$x=-\frac{7\pi}{4}$

$-\frac{5\pi}{2} \leq \frac{3\pi}{4}+2\pi m \leq -\pi;\ -10 \leq 3+8m \leq -4;\ -\frac{13}{8} \leq m \leq -\frac{7}{8}\Rightarrow m=-1;$

$x=-\frac{5\pi}{4}$

$-\frac{5\pi}{2} \leq \frac{\pi}{2}+\pi k \leq -\pi; \ -3 \leq k \leq -\frac{3}{2}; k=-3; k=-2$

$x=-\frac{5\pi}{2}; x=-\frac{3\pi}{2}$