Найдите корни уравнения, принадлежащие отрезку [π; 2π].

$\cos^{2}x = \sqrt{2}\sin 2x - \sin^{2}x \\ \\ 1 = \sqrt{2} \sin 2x \\ \\ \sin 2x = \dfrac{\sqrt{2}}{2} \\ \\ $\left[ \begin{gathered} 2x = \dfrac{\pi}{4}+ 2\pi n, n \in Z \\ 2x = \dfrac{3\pi}{4} + 2\pi k, k \in Z \\ \end{gathered} \right.$$

$\left[ \begin{gathered} x = \dfrac{\pi}{8}+ \pi n, n \in Z \\ x = \dfrac{3\pi}{8} + \pi k, k \in Z \\ \end{gathered} \right.$

Корни уравнения (1):

$\pi + \dfrac{\pi}{8} = \dfrac{9\pi}{8} \\ \\ \dfrac{3\pi}{2} - \dfrac{\pi}{8} = \dfrac{11\pi}{8}$

Ответ: 9π/8, 11π/8