3-4 cos^2x>0 как решить неравенство?

$\displaystyle 3-4cos^2x\ \textgreater \ 0$

$\displaystyle 3\ \textgreater \ 4cos^2x$

$\displaystyle cos^2x\ \textless \ \frac{3}{4}$

$\displaystyle -\frac{ \sqrt{3}}{2}\ \textless \ cosx\ \textless \ \frac{ \sqrt{3}}{2}$

1)
$\displaystyle cosx\ \textgreater \ - \frac{ \sqrt{3}}{2}$

$\displaystyle - \frac{5 \pi }{6}+2 \pi n\ \textless \ x\ \textless \ \frac{5 \pi }{6}+2 \pi n; n\in Z$

2)
$\displaystyle cosx\ \textless \ \frac{ \sqrt{3}}{2}$

$\displaystyle \frac{ \pi }{6}+2 \pi n\ \textless \ x\ \textless \ \frac{11 \pi }{6}+2 \pi n; n\in Z$

объединим промежутки

$\displaystyle (- \frac{5 \pi }{6}+2 \pi n; - \frac{\pi }{6}+2 \pi n) ( \frac{ \pi }{6}+2 \pi n; \frac{5 \pi }{6}+2 \pi n); n\in Z$