Решите уравнение \(\sin x + \cos x = 2\sqrt{2} \sin x \cos x\).

Решение

\[\sin x + \cos x = 2\sqrt{2} \sin x \cos x \Leftrightarrow\]\[\Leftrightarrow \dfrac{\sqrt{2}}{2} \sin x + \dfrac{\sqrt{2}}{2} \cos x = 2 \sin x \cos x \Leftrightarrow\]\[\Leftrightarrow \sin \left(x + \dfrac{\pi}{4}\right) = \sin 2x \Leftrightarrow\]\[\Leftrightarrow \left[\begin{array}{l} x + \dfrac{\pi}{4} = 2x + 2\pi k, k \in \mathbb{Z} \\ \\ x + \dfrac{\pi}{4} = \pi - 2x + 2\pi n, n \in \mathbb{Z} \end{array}\right. \Leftrightarrow\]\[\Leftrightarrow \left[\begin{array}{l} x = \dfrac{\pi}{4} + 2\pi k, k \in \mathbb{Z} \\ \\ x = \dfrac{\pi}{4} + \dfrac{2\pi n}{3}, n \in \mathbb{Z} \end{array}\right. \Leftrightarrow\]\[\Leftrightarrow x = \dfrac{\pi}{4} + \dfrac{2\pi n}{3}, n \in \mathbb{Z}.\]

Ответ: \(x = \dfrac{\pi}{4} + \dfrac{2\pi n}{3}, n \in \mathbb{Z}\).