Question

помогите пж ✓3tg(1/3+p/6)<1 ​

Answer 1

Ответ:

$\displaystyle \sqrt3\, tg\Big(\dfrac{x}{3}+\dfrac{\pi}{6}\Big)<1\\\\\\ tg\Big(\dfrac{x}{3}+\dfrac{\pi }{6}\Big)< \dfrac{\sqrt3}{3}\\\\\\-\frac{\pi}{2}+\pi k<\dfrac{x}{3}+\dfrac{\pi }{6}<\frac{\pi }{6}+\pi k\ \ ,\ k\in Z\\\\\\-\frac{2\pi }{3}+\pi k<\dfrac{x}{3}<\pi k\ \ ,\ \ k\in Z\\\\\\-2\pi +3\pi k<x<3\pi k\ \ ,\ k\in Z\\\\\\Otvet:\ \ x\in \Big(\ \pi (3k-2)\ ;\ 3\pi k\ \Big)\ \ ,\ k\in Z$