$\log_a f=\log_a g \Leftrightarrow \begin{cases} f=g \\ f \gt 0 / g \gt 0 \end{cases}$[NEWLINE] $\log_2(\sin 2x) = \log_2(\sqrt{2} \cos x)$[NEWLINE] $ \begin{cases} \sin 2x = \sqrt{2} \cos x \\ \sqrt{2} \cos x > 0 \end{cases} $ [NEWLINE] $ \begin{cases} 2 \sin x \cos x - \sqrt{2} \cos x=0 \\ \cos x > 0 \end{cases} $[NEWLINE] $ \begin{cases} \cos x (2 \sin x - \sqrt{2}) = 0 \\ \cos x > 0 \end{cases} $[NEWLINE] 1) $\cos x = 0 $ — не подходит [NEWLINE] 2) $2 \sin x - \sqrt{2} = 0 \Rightarrow \sin x = \dfrac{\sqrt{2}}{2}$ [NEWLINE] $ x = \dfrac{\pi}{4} + 2\pi n \quad \text{или} \quad x = \dfrac{3\pi}{4} + 2\pi n,\ n \in \mathbb{Z} $[NEWLINE] \(\)[NEWLINE] Проверим $\cos x > 0$: [NEWLINE] — $x = \dfrac{\pi}{4} + 2\pi n$: $\cos x = \dfrac{\sqrt{2}}{2} > 0$ — подходит [NEWLINE] — $x = \dfrac{3\pi}{4} + 2\pi n$: $\cos x = -\dfrac{\sqrt{2}}{2} < 0$ — не подходит[NEWLINE] \(\)[NEWLINE] Итого: $x = \dfrac{\pi}{4} + 2\pi n,\ n \in \mathbb{Z}$ [image:0:right]б) Отберем корни, принадлежащие отрезку $\left[-3\pi; -\frac{3\pi}{2}\right]$.[IMAGENEWLINE] $x_1 = -2\pi+\dfrac{\pi}{4}=-\dfrac{7\pi}{4}$[IMAGENEWLINE] [IMAGENEWLINE] [BOLD]Ответ:[/BOLD] а) $x = \dfrac{\pi}{4} + 2\pi n,\ n \in \mathbb{Z}$[IMAGENEWLINE] б) $-\dfrac{7\pi}{4}$[IMAGENEWLINE]