x = \pi n, n \in Z\\cosx = 0 => x = \frac{\pi}{2} + \pi n, n \in Z\\ cos2x = \frac{1}{2} => 2x = \pm\frac{\pi}{3} + 2\pi n. n \in Z => x = \pm\frac{\pi}{6} + \pi n, n \in Z" alt="sin2xcos2x - sinxcosx = 0\\2sinxcosxcos2x-sinxcosx=0\\2sinxcosx(cos2x-\frac{1}{2})=0\\sinx = 0 => x = \pi n, n \in Z\\cosx = 0 => x = \frac{\pi}{2} + \pi n, n \in Z\\ cos2x = \frac{1}{2} => 2x = \pm\frac{\pi}{3} + 2\pi n. n \in Z => x = \pm\frac{\pi}{6} + \pi n, n \in Z" align="absmiddle" class="latex-formula">
Ответ:
