Ответ:

Тригонометрические неравенства .

[tex]\displaystyle \bf 1)\ \ 2sinx < -1\ \ ,\ \ sinx < -\frac{1}{2}\ \ ,\\\\-\frac{5\pi }{6}+2\pi n < x < -\frac{\pi}{6}+2\pi n\ \ ,\ n\in Z\\\\\\2)\ \ -3\, tgx\geq \sqrt3\ \ ,\ \ tgx\leq -\frac{\sqrt3}{3}\ \ ,\\\\-\frac{\pi }{2}+\pi n < x\leq -\frac{\pi}{6}+\pi n\ \leq \ ,\ \ n\in Z\\\\\\3)\ \ 2\, cos\Big(3x-\frac{\pi }{3}\Big) < \sqrt3\ \ ,\ \ cos\Big(3x-\frac{\pi }{3}\Big) < \frac{\sqrt3}{2}\ \ ,\\\\\frac{\pi}{6}+2\pi n < 3x-\frac{\pi}{3} < \frac{11\pi }{6}+2\pi n\ \ ,\ \ n\in Z[/tex]

[tex]\bf \displaystyle \frac{\pi }{2}+2\pi n < 3x < \frac{13\pi }{6}+2\pi n\ \ ,\ n\in Z\\\\\frac{\pi }{6}+\frac{2\pi }{2} < x < \frac{13\pi }{18}+\frac{2\pi }{3}\ \ ,\ \ n\in Z[/tex]