Корень из 3 sin2x-cos2x =2

[latex]\sqrt3sin2x-cos2x=2|:2\\\frac{\sqrt3}{2}sin2x-\frac{1}{2}cos2x=1\\\frac{\sqrt3}{2}=cos\frac{\pi}{6}, \; \; \frac{1}{2}=sin\frac{\pi}{6};\\sin2xcos\frac{\pi}{6}-cos2xsin\frac{\pi}{6}=1\\sin(2x-\frac{\pi}{6})=1\\2x-\frac{\pi}{6}=\frac{\pi}{2}+2\pi n\\2x=\frac{2\pi}{3}+2\pi n\\x=\frac{\pi}{3}+\pi n, \; n\in Z.[/latex]