\(2\sin^3x=\sqrt2(1-\sin^2x)+2\sin x\)[NEWLINE] \(2\sin^3x-\sqrt2+\sqrt2\sin^2x-2\sin x=0\)[NEWLINE] \(\sin^2x(2\sin x+\sqrt2)-(\sqrt2+2\sin x)=0\)[NEWLINE] \((2\sin x+\sqrt2)(\sin^2x-1)=0\)[NEWLINE] [TWO-COLUMNS]\(\sin x=-\frac{\sqrt2}{2}\)[NEWLINE]\(\left[\begin{aligned} x=-\frac{\pi}{4}+2\pi n, \quad n \in Z \\ x=\frac{5\pi}{4} +2\pi k, \quad k \in Z\\\end{aligned}\right.\)[COLUMN-BREAK]\(\sin^2x-1=0\)[NEWLINE]\(\sin x=\pm1\)[NEWLINE]\(x=\frac{\pi}{2}+ \pi m, \quad m \in Z\)[/TWO-COLUMNS] б) Отберем корни с помощью тригонометрической окружности: [NEWLINE] [image:0:left] Нанесем корни и отрезок на единичную окружность.[IMAGENEWLINE] \(x_1=-\frac{5\pi}{2}\)[IMAGENEWLINE] \(x_2=-\frac{7\pi}{2}\)[IMAGENEWLINE] \(x_3=-3\pi+\frac{\pi}{4}=-\frac{11\pi}{4}.\)[NEWLINE] Ответ: а) \(\left[\begin{aligned} x=-\frac{\pi}{4}+2\pi n, \quad n \in Z \\ x=\frac{5\pi}{4} +2\pi k, \quad k \in Z \\ x=\frac{\pi}{2}+ \pi m, \quad m \in Z\\\end{aligned}\right.\)[NEWLINE] б) \(x_1=-\frac{5\pi}{2}\)[NEWLINE] \(x_2=-\frac{7\pi}{2}\)[NEWLINE] \(x_3=-3\pi+\frac{\pi}{4}=-\frac{11\pi}{4}.\)[NEWLINE]