Question

√3 tg(1/3x+π/6)<1
сделайте УМОЛЯЮ​

Answer 1

$\sqrt3\, tg(\frac{1}{3}x+\frac{\pi}{6})<1\\\\tg(\frac{1}{3}x+\frac{\pi}{6})<\frac{1}{\sqrt3}\\\\-\frac{\pi}{2}+\pi n<\frac{1}{3}x+\frac{\pi}{6}<\frac{\pi}{6}+\pi n\\\\-\frac{2\pi }{3}+\pi n<\frac{x}{3}<\pi n\\\\-2\pi +3\pi n<x<3\pi n\; ,\; n\in Z\\\\x\in (-2\pi +3\pi n\, ;\, 3\pi n)\; ,\; n\in Z$