Question

СРОЧНО!! Решите неравенство:
√‎3ctg(2x-π/4)<1

Answer 1

Ответ:

$\sqrt3\, ctg\Big(2x-\frac{\pi }{4}\Big)<1\\\\\\ctg\Big(2x-\dfrac{\pi }{4}\Big)<\dfrac{1}{\sqrt3}\\\\\\\dfrac{\pi}{3}+\pi n<2x-\dfrac{\pi }{4}<\pi +\pi n \ \ ,\ n\in Z\\\\\\\dfrac{7\pi }{12}+\pi n<2x<\dfrac{5\pi }{4}+\pi n\ \ ,\ n\in Z\\\\\\\dfrac{7\pi }{24}+\dfrac{\pi n}{2}<x<\dfrac{5\pi }{8}+\dfrac{\pi n}{2}\ \ ,\ n\in Z\\\\\\Otvet:\ \ x\in \Big(\ \dfrac{7\pi }{24}+\dfrac{\pi n}{2}\ ;\ \dfrac{5\pi }{8}+\dfrac{\pi n}{2}\ \Big)\ ,\ n\in Z\ .$