Решите неравенство √3 sin2x+cos2x= -√2

Решите задачу:

$\sqrt3sin2x+cos2x=-\sqrt2\, |:2\\\\\frac{\sqrt3}{2}sin2x+\frac{1}{2}cos2x=-\frac{\sqrt2}{2}\\\\cos\frac{\pi}{6}sin2x+sin\frac{\pi}{6}cos2x=-\frac{\sqrt2}{2}\\\\sin(2x+\frac{\pi}{6})=-\frac{\sqrt2}{2}\\\\2x+\frac{\pi}{6}=(-1)^{n}\cdot (-\frac{\pi}{4})+\pi n=(-1)^{n+1}\cdot \frac{\pi}{4}+\pi n\; ,\; n\in Z\\\\2x=(-1)^{n+1}\cdot \frac{\pi}{4}-\frac{\pi}{6} +\pi n,\; n\in Z\\\\x=(-1)^{n+1}\frac{\pi}{8}-\frac{\pi}{12}+\frac{\pi n}{2},\; n\in Z$