\[ \cos\left(2x + \frac{\pi}{2}\right) = -\sin 2x \] [NEWLINE] \[ \sin 2x = 2 \sin x \cos x \] [NEWLINE] \(\)[NEWLINE] \[ 4\sqrt{3} \cos^3 x = -2 \sin x \cos x \] [NEWLINE] \[ 4\sqrt{3} \cos^3 x + 2 \sin x \cos x = 0 \] [NEWLINE] \[ 2 \cos x (2\sqrt{3} \cos^2 x + \sin x) = 0 \] [NEWLINE] [TWO-COLUMNS]\(\cos x = 0\) [NEWLINE]\[ x = \frac{\pi}{2} + \pi m, \quad m \in \mathbb{Z} \] [NEWLINE][COLUMN-BREAK]\(2\sqrt{3} \cos^2 x + \sin x = 0\) [NEWLINE]\[ 2\sqrt{3} (1 - \sin^2 x) + \sin x = 0 \] [NEWLINE]\[ 2\sqrt{3} - 2\sqrt{3} \sin^2 x + \sin x = 0 \] [NEWLINE] \[ 2\sqrt{3} \sin^2 x - \sin x - 2\sqrt{3} = 0 \][NEWLINE] \(t = \sin x,\quad |t| \ leq 1\): [NEWLINE]\[ 2\sqrt{3} t^2 - t - 2\sqrt{3} = 0 \] [NEWLINE]\[ D = 1 + 48 = 49 \] [NEWLINE]\[ t_1 = \frac{1 + 7}{4\sqrt{3}} = \frac{2}{\sqrt{3}} \] (не подходит, так как \(|t| \leq 1\)) [NEWLINE]\[ t_2 = \frac{1 - 7}{4\sqrt{3}} = -\frac{\sqrt{3}}{2} \] [NEWLINE] \[x = \left[\begin{aligned}& -\frac{\pi}{3} + 2\pi n \\& \frac{4\pi}{3} + 2\pi k \\\end{aligned}\right., \quad n,k \in \mathbb{Z} \] б) Проведем отбор корней с помощью тригонометрической окружности [NEWLINE] [image:0]$x_1, x_2 $ получаются при пересечении окружности и плоскости, $x_3$ получается движением от $-3\pi$ вперёд на $\frac{\pi}{3} \rightarrow x_1=-\frac{7\pi}{2},$[IMAGENEWLINE]$ x_2=-\frac{5\pi}{2}, $ [IMAGENEWLINE]$x_3=-3\pi+\frac{\pi}{3}=-\frac{8\pi}{3}.$[IMAGENEWLINE][IMAGENEWLINE] [BOLD]Ответ: [\BOLD] а) \[x = \left[\begin{aligned}& -\frac{\pi}{3} + 2\pi n \\& \frac{4\pi}{3} + 2\pi k & x = \frac{\pi}{2} + \pi n, \quad n \in \mathbb{Z}\\\end{aligned}\right., \quad n,k,m \in \mathbb{Z} \][IMAGENEWLINE] б) $x_1=-\frac{7\pi}{2},$[IMAGENEWLINE]$ x_2=-\frac{5\pi}{2}, $ [IMAGENEWLINE]$x_3=-\frac{8\pi}{3}.$