Điều kiện xác định $\cos2x\ne 0, \sin 2x\ne 0, \sin 3x\ne 0$
$ \Leftrightarrow \left\{ \begin{array}{l} 2x \ne k\pi \\ 2x \ne \dfrac{\pi }{2} + k\pi \\ 3x \ne k2\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne k\dfrac{\pi} 2 \\ x \ne \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\ x \ne \dfrac{{k2\pi }}{3} \end{array} \right.$
$\begin{array}{l} \dfrac{{\sin 3x}}{{\cos 2x}} + \dfrac{{\cos 3x}}{{\sin 2x}} = \dfrac{2}{{\sin 3x}}\\ \Leftrightarrow \dfrac{{\sin 3x\sin 2x + \cos 3x\cos 2x}}{{\cos 2x\sin 2x}} = \dfrac{2}{{\sin 3x}}\\ \Leftrightarrow \dfrac{{\dfrac{1}{2}\left( {\cos x - \cos 5x} \right) + \dfrac{1}{2}\left( {\cos x + \cos 5x} \right)}}{{\dfrac{1}{2}.\sin 4x}} = \dfrac{2}{{\sin 3x}}\\ \Leftrightarrow \dfrac{{2\cos x}}{{\sin 4x}} = \dfrac{2}{{\sin 3x}}\\ \Leftrightarrow 2\cos x\sin 3x = 2\sin 4x\\ \Leftrightarrow \left( {\sin 4x + \sin 2x} \right) = 2\sin 4x\\ \Leftrightarrow \sin 2x = \sin 4x\\ \Leftrightarrow \left[ \begin{array}{l} 4x = 2x + k2\pi \\ 4x = \pi - 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi (L)\\ x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3} \end{array} \right.\\ \Rightarrow x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3} \end{array}$
