Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{6} + k\pi\\x = -\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\dfrac{3}{\sin^2x} - 2\sqrt3\cot x - 6 = 0\qquad (*)$
$ĐK: \sin x \ne 0 \Leftrightarrow x \ne n\pi$
$(*)\Leftrightarrow 3(\cot^2x +1) - 2\sqrt3\cot x - 6 = 0$
$\Leftrightarrow 3\cot^2x -2\sqrt3\cot x - 3 = 0$
$\Leftrightarrow \left[\begin{array}{l}\cot x =\sqrt3\\\cot x = -\dfrac{\sqrt3}{3}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{6} + k\pi\\x = -\dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
