Question

tg4x > -1/√3.
Тригонаметрия

Answer 1

$tg4x>-\frac{1}{\sqrt{3} }\\\\arctg(-\frac{1}{\sqrt{3}})+\pi n <4x<\frac{\pi }{2} +\pi n,n\in Z \\\\-\frac{\pi }{6} +\pi n <4x<\frac{\pi }{2} +\pi n,n\in Z\\\\\boxed{-\frac{\pi }{24}+\frac{\pi n }{4} <x<\frac{\pi }{8}+\frac{\pi n }{4}.n\in Z}$