Đáp án:

\(\left[ \begin{array}{l} x = - \frac{\pi }{6} + k2\pi \\ x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\)

Giải thích các bước giải: \(\begin{array}{l} \frac{{3\left( {\cos x + \cot x} \right)}}{{\cot x - \cos x}} - 2\sin x = 2\\ DK:\,\,\,\left\{ \begin{array}{l} \sin x \ne 0\\ \cot x - \cos x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \sin x \ne 0\\ \cos x\left( {1 - \sin x} \right) \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \sin x \ne 0\\ \cos x \ne 0 \end{array} \right. \Leftrightarrow \sin 2x \ne 0\\ pt \Leftrightarrow \frac{{3\left( {\cos x + \cot x} \right)}}{{\cot x - \cos x}} = 2 + 2\sin x\\ \Leftrightarrow 3\left( {\cos x + \cot x} \right) = \left( {2\sin x + 2} \right)\left( {\cot x - \cos x} \right)\\ \Leftrightarrow 3\left( {\cos x + \frac{{\cos x}}{{\sin x}}} \right) = \left( {2\sin x + 2} \right)\left( {\frac{{\cos x}}{{\sin x}} - \cos x} \right)\\ \Leftrightarrow 3\left( {\sin x.\cos x + \cos x} \right) = \left( {2\sin x + 2} \right)\left( {\cos x - \sin x\cos x} \right)\\ \Leftrightarrow 3\cos x\left( {\sin x + 1} \right) = 2\left( {\sin x + 1} \right)\cos x\left( {1 - \sin x} \right)\\ \Leftrightarrow \cos x\left[ {3\left( {\sin x + 1} \right) - 2\left( {\sin x + 1} \right)\left( {1 - \sin x} \right)} \right] = 0\\ \Leftrightarrow 3\left( {\sin x + 1} \right) - 2\left( {\sin x + 1} \right)\left( {1 - \sin x} \right) = 0\,\,\,\,\left( {do\,\,\,\cos x \ne 0} \right)\\ \Leftrightarrow \left( {\sin x + 1} \right)\left( {3 - 2 + 2\sin x} \right) = 0\\ \Leftrightarrow \left( {\sin x + 1} \right)\left( {1 + 2\sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = - 1\,\,\,\left( {ktm\,\,\,do\,\,\cos x \ne 0} \right)\\ \sin x = - \frac{1}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - \frac{\pi }{6} + k2\pi \\ x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\)