Question

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

Answer 1

$1. \ tg \bigg (x - \dfrac{\pi }{3} \bigg )= \sqrt{3} \\ \\ x - \dfrac{\pi }{3} = \dfrac{\pi }{3} + \pi n, \ n \in Z \\ \\ \boxed{x = \dfrac{2 \pi }{3} + \pi n, \ n \in Z}$

$2. \ cosx \ \textgreater \ - \dfrac{ \sqrt{3} }{2} \\ \\ -arccos(- \dfrac{ \sqrt{3} }{2}) + 2 \pi n \ \textless \ x \ \textless \ arccos(- \dfrac{ \sqrt{3} }{2}) + 2 \pi n, \ n \in Z \\ \\ -\dfrac{5 \pi }{6} + 2 \pi n \ \textless \ x \ \textless \ \dfrac{5 \pi }{6} +2 \pi n, \ n \in Z$

Answer 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
$cosx\ \textgreater \ -\dfrac{\sqrt3}{2}\\-arccos(-\dfrac{\sqrt3}{2})\ \textless \ x\ \textless \ arccos(-\dfrac{\sqrt3}{2})\\-\pi-\dfrac{\pi}{6}+2\pi k\ \textless \ x\ \textless \ \pi-\dfrac{\pi}{6}+2\pi k,k\in\mathbb{Z}}\\\boxed{-\dfrac{5\pi}{6}+2\pi k\ \textless \ x\ \textless \ \dfrac{5\pi}{6}+2\pi k,k\in\mathbb{Z}}$