Sin2x-cos2x=√3/2 Помогите пожалуйста

$sin2x-cos2x=\frac{\sqrt3}2$

Воспользуемся такой формулой: $sinx-cosx=\sqrt2sin(x - \frac{\pi}4)$

$\sqrt{2}sin(2x-\frac{\pi}4)=\frac{\sqrt3}2\\sin(2x-\frac{\pi}4)=\frac{\sqrt3}{2\sqrt2}\\sin(2x-\frac{\pi}4)=\frac{\sqrt3*\sqrt2}{2\sqrt2*\sqrt2}\\sin(2x-\frac{\pi}4)=\frac{\sqrt6}{4}\\2x-\frac{\pi}4=arcsin(\frac{\sqrt6}4)+2\pi n;n\in Z\\2x-\frac{\pi}4 = \pi - arcsin(\frac{\sqrt6}4)+2\pi n;n\in Z\\2x=\frac{\pi}4+arcsin(\frac{\sqrt6}4)+2\pi n;n\in Z\\2x = \frac{5\pi}4 - arcsin(\frac{\sqrt6}4)+2\pi n;n\in Z$
$x=\frac{\pi}8+\frac{1}2arcsin(\frac{\sqrt6}4)+\pi n;n\in Z\\x = \frac{5\pi}8 - \frac{1}2arcsin(\frac{\sqrt6}4)+\pi n;n\in Z$