Question

(sin^2x-cos^2x)^2-4cos^2xsin^2x=-0,5

Answer 1

$(sin^2x-cos^2x)^2-4cos^2xsin^2x=-1/2\\\\cos^22x-sin^22x=-1/2\\\\cos(4x)=-1/2\\\\4x=\pm arccos(-1/2)+2\pi n, n\in Z\\\\x=\pm 1/4 arccos(-1/2)+\pi/2 n, n\in Z$

$arccos(-1/2)= \pi-arccos(1/2)=\pi-\pi/3=2\pi/3$

Значит, $x=\pm \frac{\pi}{6} + \frac{\pi n}{2}, n\in Z$