Алгебра: sin(-x)*sinx=cosx

$\sqrt{2}sin(\frac{2\pi}2-x)*sinx=cosx$

Воспользуемся формулой приведения.
$sin(\pi-x)=sinx$

$\sqrt{2}*sinx*sinx=cosx\\\sqrt{2}sin^2x-cosx=0\\\sqrt{2}(1-cos^2x)-cosx=0\\-\sqrt{2}(cos^2x-1)-cosx=0\\\sqrt{2}cos^2x-\sqrt{2}+cosx=0\\$
Пусть:
$t=cosx;\,\,t\in[-1;1]\\t^2\sqrt{2}+t-\sqrt{2}=0\\D=1+4*\sqrt{2}*\sqrt{2}=1+8=9\\\\t_1=\frac{-1+3}{2\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt2}2\\\\t_2=\frac{-1-3}{2\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}\,\,\notin t\\\\cosx=\frac{\sqrt2}2\\x=б\frac{\pi}4+2\pi n;n\in Z$