Question

Решите неравенства
Cos x > корень 3:2
Sin 2x> - корень 2:2
Cos 2 x < корень 2:2

Answer 1

Ответ:

$1)\ \ cosx>\dfrac{\sqrt3}{2}\ \ ,\ \ x\in \Big(-\dfrac{\pi}{6}+2\pi n\ ;\ \dfrac{\pi }{6}+2\pi n\Big)\ ,\ n\in Z\\\\\\2)\ \ sin2x>-\dfrac{\sqrt2}{2}\ \ ,\ \ \ -\dfrac{\pi}{4}+2\pi n<2x<\dfrac{5\pi }{4}+2\pi n\ ,\ n\in Z\ ,\\\\\\x\in \Big(-\dfrac{\pi}{8}+\pi n\ ;\ \dfrac{5\pi }{8}+\pi n\Big)\ ,\ n\in Z\\\\\\3)\ \ cos2x<\dfrac{\sqrt2}{2}\ \ ,\ \ \ \ \dfrac{\pi}{4}+2\pi n<2x<\dfrac{7\pi}{4}+2\pi n\ ,\ n\in Z\\\\\\x\in \Big(\ \dfrac{\pi}{8}+\pi n\ ;\ \dfrac{7\pi}{8}+\pi n\Big)\ ,\ n\in Z$