Đáp án + Giải thích các bước giải:

ĐKXĐ: $x \ne \dfrac{2\pi}{3} + k\pi; \dfrac{\pi}{2} + k\pi(k \in \mathbb{Z})$ 

$\dfrac{\sin 2x + 2 \cos x - \sin x - 1}{\tan x + \sqrt{3}} = 0$
$\Leftrightarrow \sin 2x + 2 \cos x - \sin x - 1 = 0$

$\Leftrightarrow 2 \sin x \cos x + 2 \cos x - \sin x - 1 = 0$

$\Leftrightarrow 2 \cos x(\sin x + 1) - (\sin x + 1) = 0$
$\Leftrightarrow (2 \cos x - 1)(\sin x + 1) = 0$

$\Leftrightarrow$ $\left[ \begin{array}{l}2\cos x - 1 =0\\\sin x + 1 =0\end{array} \right.$

$\Leftrightarrow$ $\left[ \begin{array}{l}\cos x =\dfrac{1}{2}\\\sin x =-1\end{array} \right.$ 

$\Leftrightarrow$ $\left[ \begin{array}{l} x = \dfrac{\pi}{3} + k2\pi (tm) \\ x = -\dfrac{\pi}{3} + k2\pi (ktm) \\ x = -\dfrac{\pi}{2} + k2\pi (ktm) \end{array} (k \in \mathbb{Z}) \right.$

Vậy $S =$`{\pi/3 + k2\pi | k \in ZZ}`