\sqrt{3} tg(\frac{\pi}{3}-x)-1\leq 0

Решите задачу:

$\displaystyle \sqrt{3} tg(\frac{\pi}{3}-x)-1\leq 0\\\\tg( \frac{ \pi }{3}-x) \leq \frac{1}{ \sqrt{3}}\\\\ * - \frac{ \pi }{2} \ \textless \ \frac{ \pi }{3}-x \leq \frac{ \pi }{6}; (+2 \pi n; n\in Z)\\\\ **\frac{ \pi }{2} \ \textless \ \frac{ \pi }{3}-x \leq \frac{7 \pi }{6}; (+2 \pi n; n\in Z)\\\\ * - \frac{ \pi }{2}- \frac{ \pi }{3} \ \textless \ -x \leq \frac{ \pi }{6}- \frac{ \pi }{3}; (+2 \pi n; n\in Z)\\\\ ** \frac{ \pi }{2}- \frac{ \pi }{3} \ \textless \ -x \leq \frac{7 \pi }{6}- \frac{ \pi }{3}; (+2 \pi n; n\in Z)$

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