а) $\sin(x - \pi) = -\sin x$[NEWLINE] \(\)[NEWLINE] $\sin 2x - \sqrt{3} \sin x = 0$[NEWLINE] $2 \sin x \cos x - \sqrt{3} \sin x = 0$[NEWLINE] $\sin x (2 \cos x - \sqrt{3}) = 0$[NEWLINE] [TWO-COLUMNS]$\sin x = 0$[NEWLINE]$x = \pi n, \quad n \in \mathbb{Z}$[COLUMN-BREAK]$2 \cos x - \sqrt{3} = 0$[NEWLINE]$\cos x = \frac{\sqrt{3}}{2}$[NEWLINE]$x = \pm \frac{\pi}{6} + 2\pi k, \quad k \in \mathbb{Z}$[/TWO-COLUMNS] [image:0:right]б) Отберем корни, принадлежащие отрезку $\left[-\frac{7\pi}{2}; -2\pi\right]$.[IMAGENEWLINE] $x_1 = -3\pi$[IMAGENEWLINE] $x_2 = -2\pi$[IMAGENEWLINE] $x_3 = -\frac{\pi}{6} - 2\pi = -\frac{13\pi}{6}$[IMAGENEWLINE][IMAGENEWLINE] [BOLD]Ответ:[/BOLD] а) $x = \pi n, \ n \in \mathbb{Z}, \ x = \pm \frac{\pi}{6} + 2\pi k, \ k \in \mathbb{Z}$[IMAGENEWLINE] б) $-3\pi, -2\pi, -\frac{13\pi}{6}$[IMAGENEWLINE]