Question

2sin3x
$> - \sqrt{2}$
​

Answer 1

$\sin3x>-\frac{\sqrt{2}}{2}$

Тогда:

$-\frac{\pi}{4}+\pi k<3x<\frac{5\pi}{4}+\pi k$, k=0,1,2...

Итого:

$-\frac{\pi}{12}+\frac{\pi k}{3}<x<\frac{5\pi}{12}+\frac{\pi k}{3}$