Question

Решите неравенство cos 2x > -корень из 3/2

Answer 1

$- \frac{5 \pi }{6}+2 \pi k<2x< \frac{5 \pi }{6}+2 \pi k,k\in Z \\ - \frac{5 \pi }{12}+ \pi k<x<\frac{5 \pi }{12}+\pi k,k\in Z$

Answer 2

$\cos2x>-\frac{\sqrt3}{2};\\ \cos2x=-\frac{\sqrt3}{2};\\ \cos2x=\pm\frac{5\pi}{6}+2\pi n,\n\in Z;\\ -\frac{5\pi}{6}+2\pi n<2x<\frac{5\pi}{6}+2\pi n;\\ -\frac{5\pi}{12}+\pi n<x<\frac{5\pi}{12}+\pi n;\\ \\ \\ x\in\left(-\frac{5\pi}{12}+\pi n;\frac{5\pi}{12}+\pi n\right),\ n\in Z.$