0 \\ -\cos 2x>0 \\ \cos 2x<0 \\ \frac{\pi}{2} + \pi n < 2x < \frac{3\pi}{2} + \pi n|:2 \\ \frac{\pi}{4} + \frac{\pi n}{2} < x < \frac{3\pi }{4} + \frac{\pi n}{2} " alt="\cos(2x- \pi )>0 \\ -\cos 2x>0 \\ \cos 2x<0 \\ \frac{\pi}{2} + \pi n < 2x < \frac{3\pi}{2} + \pi n|:2 \\ \frac{\pi}{4} + \frac{\pi n}{2} < x < \frac{3\pi }{4} + \frac{\pi n}{2} " align="absmiddle" class="latex-formula">
Ответ:
