Question

Решить неравенство:
cos(pi/8)*cos(x)-sin(x)*sin(pi/8)< - корень из трёх, делённое на 2

Answer 1

Ответ:

$\displaystyle cos\frac{\pi}{8}\cdot cosx-sinx\cdot sin\frac{\pi}{8}<-\frac{\sqrt3}{2}\\\\\\cos\Big(x+\frac{\pi}{8}\Big)<-\frac{\sqrt3}{2}\\\\\\\frac{5\pi }{6}+2\pi n<x+\frac{\pi}{8}<\frac{7\pi}{6}+2\pi n\ ,\ n\in Z\\\\\\\frac{5\pi }{6}-\frac{\pi}{8}+2\pi n<x<\frac{7\pi}{6}-\frac{\pi}{8}+2\pi n\ ,\ n\in Z\\\\\\\frac{17\pi }{24}+2\pi n\ <x<\ \frac{25\pi}{24}+2\pi n\ ,\ n\in Z\\\\\\x\in \Big(\ \frac{17\pi }{24}+2\pi n\ ;\ \frac{25\pi}{24}+2\pi n\ \Big)\ ,\ n\in Z$