Question

Решите неравенство
√‎3ctg(2x-π/4)>1

Answer 1

Ответ.

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