\(cos(\pi+2x)=-cos2x\), так как это третья четверть; \(sin^2x=1-cos^2x\); \(cos2x=cos^2x-sin^2x\).[NEWLINE] \(2+2cos2x-\sqrt8cosx-\sqrt6+\sqrt{12}cosx=0\). Заменяем cos2x [NEWLINE] \(2+2(cos^2x-sin^2x)-\sqrt8cosx-\sqrt6+\sqrt{12}cosx=0\) [NEWLINE] \(2+2(cos^2x-1+cos^2x)-\sqrt8cosx-\sqrt6+\sqrt{12}cosx=0\) [NEWLINE] \(2+2cos^2x-2+2cos^2x-\sqrt8cosx-\sqrt6+\sqrt{12}cosx=0\) [NEWLINE] \(2+2cos^2x-2+2cos^2x-\sqrt8cosx-\sqrt6+\sqrt{12}cosx=0\) [NEWLINE] \(4cos^2x-2\sqrt2cosx-\sqrt3\sqrt2+2\sqrt{3}cosx=0\) [NEWLINE] \(2cosx(2cosx-\sqrt2)+\sqrt3(-\sqrt2+2cosx)=0\) [NEWLINE] \((2cosx+\sqrt3)(2cosx-\sqrt2)=0\) [NEWLINE] [TWO-COLUMNS]\(cosx=-\frac{\sqrt3}2{}\)[NEWLINE]\(x=\pm\frac{5\pi}{6}+2\pi n,n \in Z\)[COLUMN-BREAK]\(cosx=\frac{\sqrt2}{2}\)[NEWLINE]\(x=\pm\frac{\pi}{4}+2\pi n,n \in Z\)[/TWO-COLUMNS] б) Отберем корни с помощью тригонометрической окружности: [NEWLINE] [image:0:left] Нанесем корни и отрезок на единичную окружность.[IMAGENEWLINE] \(x_1=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\) [IMAGENEWLINE] \(x_2=\pi+\frac{\pi}{6}=\frac{7\pi}{6}\) [IMAGENEWLINE] \(x_3=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}\). [IMAGENEWLINE] Ответ: а) \(x=\pm\frac{5\pi}{6}+2\pi n,n \in Z\); \(x=\pm\frac{\pi}{4}+2\pi n,n \in Z\) [IMAGENEWLINE] б) \(x_1=\frac{5\pi}{6}\); \(x_2=\frac{7\pi}{6}\); \(x_3=\frac{7\pi}{4}\)