Решите неравенство:4соs^2x-3>0

Решите задачу:

$4cos^2x-3\ \textgreater \ 0\\\\cos^2x\ \textgreater \ \frac{3}{4}\\\\\frac{1+cos2x}{2}\ \textgreater \ \frac{3}{4}\\\\1+cos2x\ \textgreater \ \frac{3}{2}\\\\cos2x\ \textgreater \ \frac{1}{2}\\\\-\frac{\pi}{3}+2\pi n\ \textless \ 2x\ \textless \ \frac{\pi}{3}+2\pi n\; ,\; n\in Z\\\\-\frac{\pi}{6}+\pi n\ \textless \ x\ \textless \ \frac{\pi}{6}+\pi n\; ,\; n\in Z\\\\P.S.\quad Formyla:\; \; \; cos^2x=\frac{1+cos2x}{2}$