Question

tg2x≥-1 можна с решением​

Answer 1

Ответ:

$tg2x\geq -1\\\\\\-\dfrac{\pi}{4}+\pi n\leq 2x<\dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\\\-\dfrac{\pi}{8}+\dfrac{\pi n}{2}\leq x<\dfrac{\pi}{4}+\dfrac{\pi n}{2}\ ,\ n\in Z\\\\\\x\in \Big[-\dfrac{\pi}{8}+\dfrac{\pi n}{2}\ ;\ \dfrac{\pi}{4}+\dfrac{\pi n}{2}\ \Big)\ ,\ n\in Z$