Пожалуйста помогите) 1. tg(x-π/3)=√3 2. cos x> -√3 /2

№1
$tg(x- \dfrac{\pi}{3})=\sqrt3\\ x-\dfrac{\pi}{3}= arctg(\sqrt3)+\pi k,k\in\mathbb{Z}\\x-\dfrac{\pi}{3}= \frac{\pi}{3}+\pi k,k\in\mathbb{Z}\\\boxed{x= \dfrac{2\pi}{3} +\pi k,k\in\mathbb{Z}}$
№2
$cosx\ \textgreater \ -\dfrac{\sqrt3}{2}\\-arccos(-\dfrac{\sqrt3}{2})\ \textless \ x\ \textless \ arccos(-\dfrac{\sqrt3}{2})\\-\pi-\dfrac{\pi}{6}+2\pi k\ \textless \ x\ \textless \ \pi-\dfrac{\pi}{6}+2\pi k,k\in\mathbb{Z}}\\\boxed{-\dfrac{5\pi}{6}+2\pi k\ \textless \ x\ \textless \ \dfrac{5\pi}{6}+2\pi k,k\in\mathbb{Z}}$