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Пункт а)

$2\cos^{3}x+\sqrt{3}\cos^{2}x+2\cos x+\sqrt{3}=0$ $cos^{2}x\,(2\cos x+\sqrt{3})+(2\cos x+\sqrt{3})=0 \;\Longleftrightarrow\; (2\cos x+\sqrt{3})(\cos^{2}x+1)=0 \;\Longleftrightarrow\; \begin{cases} \cos x=-\dfrac{\sqrt{3}}{2},\\[4pt] \cos^{2}x=-1. \end{cases}$

Уравнение $\cos^{2}x=-1$ решений не имеет, значит $\cos x=-\dfrac{\sqrt{3}}{2}$ .

Тогда $x=\pm\dfrac{5\pi}{6}+2\pi k,\quad k\in\mathbb{Z}.$

Пункт б) Отберём корни, принадлежащие отрезку $\left[-2\pi;\,-\dfrac{\pi}{2}\right]$

$x=-\dfrac{7\pi}{6},\; -\dfrac{5\pi}{6}.$

Ответ: а) $x=\pm\dfrac{5\pi}{6}+2\pi k,\; k\in\mathbb{Z}$

б) $-\dfrac{7\pi}{6},\ -\dfrac{5\pi}{6}$

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Задание №23691 ЕГЭ по Математике