Đáp án:
$x = \dfrac{5\pi}{6} + k2\pi\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\dfrac{\sqrt3\sin2x +6\sin x + \cos2x - 5}{\cos x }=2\sqrt3\quad (*)$
$ĐK:\, x \ne \dfrac{\pi}{2} + n\pi$
$(*)\Leftrightarrow 2\sqrt3\sin x\cos x + 6\sin x + 1 - 2\sin^2x - 5 = 2\sqrt3\cos x$
$\Leftrightarrow (2\sqrt3\sin x\cos x - 2\sqrt3\cos x) - 2(\sin^2x - 3\sin x + 2) = 0$
$\Leftrightarrow 2\sqrt3\cos x(\sin x - 1) - 2(\sin x -1)(\sin x -2) = 0$
$\Leftrightarrow (\sin x -1)(\sqrt3\cos x - \sin x + 2) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = 1\quad (*)\\\sin x -\sqrt3\cos x = 2 \quad (**)\end{array}\right.$
$(*)\Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\quad (loại)$
$(**)\Leftrightarrow \dfrac{1}{2}\sin x - \dfrac{\sqrt3}{2}\cos x = 1$
$\Leftrightarrow \sin\left(x - \dfrac{\pi}{3}\right) = 1$
$\Leftrightarrow x - \dfrac{\pi}{3} = \dfrac{\pi}{2} + k2\pi$
$\Leftrightarrow x = \dfrac{5\pi}{6} + k2\pi\quad (k\in\Bbb Z)$
Vậy phương trình có họ nghiệm là $x = \dfrac{5\pi}{6} + k2\pi\quad (k\in\Bbb Z)$
