Question

Решение тригон неравенств

Answer 1

$Sin(2x+\frac{\pi }{6} )>\frac{1}{2}\\\\arc Sin\frac{1}{2}+2\pi n< 2x+\frac{\pi }{6}<\pi-arc Sin\frac{1}{2}+2\pi n,n\in Z\\\\\frac{\pi }{6} +2\pi n<2x+\frac{\pi }{6}<\pi-\frac{\pi }{6}+2\pi n,n\in Z\\\\\frac{\pi }{6}+2\pi n<2x+\frac{\pi }{6}<\frac{5\pi }{6}+2\pi n,n\in Z\\\\\frac{\pi }{6}-\frac{\pi }{6}+2\pi n <2x+\frac{\pi }{6}-\frac{\pi}{6}<\frac{5\pi}{6}-\frac{\pi }{6}+2\pi n,n\in Z\\\\2\pi n<2x<\frac{2\pi }{3}+2\pi n,n\in Z\\\\\pi n<x<\frac{\pi }{3}+\pi n,n\in Z$

$Otvet:\boxed{x\in(\pi n;\frac{\pi }{3}+\pi n),n\in Z}$