Question

Решить неравенство 3ctg(π/6+x/2)>-√3

Answer 1

$3ctg (\frac{ \pi }{6}+ \frac{x}{2})>- \sqrt{3};$
$ctg (\frac{ \pi }{6}+ \frac{x}{2})>- \frac{ \sqrt{3}}{3};$
$arcctg(- \frac{ \sqrt{3}}{3})= \pi -arcctg\frac{ \sqrt{3}}{3}= \pi - \frac{ \pi }{3}= \frac{2 \pi }{3};$
$ctgt>- \frac{ \sqrt{3}}{3};$
$0+\pi n<t<\frac{2 \pi }{3}+\pi n,n \in Z;$
$0+\pi n<\frac{ \pi }{6}+ \frac{x}{2}<\frac{2 \pi }{3}+\pi n,n \in Z;$
$-\frac{ \pi }{6}+\pi n< \frac{x}{2}<\frac{2 \pi }{3}-\frac{ \pi }{6}+\pi n,n \in Z;$
$-\frac{ \pi }{3}}+2 \pi n<x< \pi +2 \pi n,n \in Z;$