Решить уравнение Tg^2(2x-Пи/6)=3

Решите задачу:

$tg^2(2x-\frac{\pi}{6})=3\\\\tg^2(2x-\frac{\pi}{6})-(\sqrt3)^2=0\\\\(tg(2x-\frac{\pi}{6})-\sqrt3)(tg(2x-\frac{\pi}{6})+\sqrt3)=0\\\\1.\; tg(2x-\frac{\pi}{6})=\sqrt3\\\\2x-\frac{\pi}{6}=\frac{\pi}{3}+\pi n,\\\\2x=\frac{\pi}{2}+\pi n,\\\\x=\frac{\pi}{4}+\frac{\pi n}{2},n\in Z\\\\2.\; tg(2x-\frac{\pi}{6})=-\sqrt3\\\\$

$2x-\frac{\pi}{6}=-\frac{\pi}{3}+\pi k,\\\\x=-\frac{\pi}{12}+\frac{\pi k }{2},k\in Z$