2cosx- sinx>0

Решите задачу:

$2cosx-\sqrt3sinx\ \textgreater \ 0|:2\\\frac{1}{2}cosx-\frac{\sqrt3}{2}sinx\ \textgreater \ 0\\\frac{1}{2}=sin\frac{\pi}{6}, \; \frac{\sqrt3}{2}=cos\frac{\pi}{6};\\sin\frac{\pi}{6}cosx-cos\frac{\pi}{6}sinx\ \textgreater \ 0\\-sin(x-\frac{\pi}{6})\ \textgreater \ 0\\sin(x-\frac{\pi}{6})\ \textless \ 0\\ -\pi+2\pi n\ \textless \ x-\frac{\pi}{6}\ \textless \ 2\pi n\\-\frac{5\pi}{6}+2\pi n\ \textless \ x\ \textless \ 2\pi +\frac{\pi}{6}, \; n\in Z$