б) \(2\sin x-2\sqrt2\sin x-4\cos^2x-\sqrt2+4=0\)[NEWLINE] \(2\sin x-2\sqrt2\sin x-4+4\sin^2x-\sqrt2+4=0\)[NEWLINE] \(2\sin x-2\sqrt2\sin x+4\sin^2 x-\sqrt2=0\)[NEWLINE] \(2\sin x(1+2\sin x)-\sqrt2(2\sin x+1)=0\)[NEWLINE] \((2\sin x+1)(2\sin x-\sqrt2)=0\)[NEWLINE] [TWO-COLUMNS]\(2\sin x+1=0\)[NEWLINE]\(\sin x=-\frac{1}{2}\)[NEWLINE]\(\left[\begin{aligned} & x=-\frac{\pi}{6}+2\pi m, \quad n \in Z \\& x=\frac{7\pi}{6}+2\pi j, \quad j \in Z \\\end{aligned}\right.\)[COLUMN-BREAK]\(2\sin x-\sqrt2=0\)[NEWLINE]\(\sin x=\frac{\sqrt2}{2}\)[NEWLINE]\(\left[\begin{aligned} & x=\frac{\pi}{4}+2\pi n, \quad n \in Z \\& x=\frac{3\pi}{4}+2\pi k, \quad k \in Z \\\end{aligned}\right.\)[/TWO-COLUMNS] Проведем отбор корней с помощью тригонометрической окружности. [NEWLINE] [image:0:left] Идем от нуля до необходимых корней: [IMAGENEWLINE] \(x_1=0-\frac{\pi}{6}=-\frac{\pi}{6}\)[IMAGENEWLINE] \(x_2=0+\frac{\pi}{4}=\frac{\pi}{4}\)[IMAGENEWLINE] Конечно, мы так же можем дойти и до корня во второй четверти, но для учебных целей можно пойти и от \(\pi\) (см.рис.):[IMAGENEWLINE] \(x_3=\pi-\frac{\pi}{4}=\frac{3\pi}{4}\).[NEWLINE] [BOLD]Ответ:[/BOLD] а) \(x = \frac{\pi}{4} + 2\pi n\); \(x=\frac{3\pi}{4} + 2\pi k\); \(x = -\frac{\pi}{6} + 2\pi m\); \(x = \frac{7\pi}{6} + 2\pi j\)[NEWLINE] б)\(x_1=-\frac{\pi}{6}\);[NEWLINE] \(x_2=\frac{\pi}{4}\);[NEWLINE] \(x_3=\frac{3\pi}{4}\).