Question

Решить тригонометрические неравенства. Даю 50 баллов
$\sqrt{3} tgx\leq 3$
$2cosx-1<0$

Answer 1

Ответ:

$\displaystyle 1)\ \ \sqrt3\, tgx\leq 3\ \ \ \Rightarrow\ \ \ \ tgx\leq \frac{\sqrt3}{3}\\\\\\-\frac{\pi}{2}+\pi n\, <\, x\, \leq \frac{\pi}{6}+\pi n\ \ ,\ n\in Z\\\\\\x\in \Big(\, -\frac{\pi}{2}+\pi n\ ;\ \frac{\pi}{6}+\pi n\ \Big],\ n\in Z$

$\displaystyle 2)\ \ 2\, cosx-1<0\ \ \ \Rightarrow\ \ \ \ cosx<\frac{1}{2}\\\\\\\frac{\pi}{3}+2\pi n<x<\frac{5\pi}{3}+2\pi n\ \ ,\ n\in Z\\\\\\x\in \Big(\ \frac{\pi}{3}+2\pi n\ ;\ \frac{5\pi}{3}+2\pi n\ \Big)\ ,\ n\in Z$