$\begin{array}{l} \sin 2x + \cos 2x + 3\sin x - \cos x - 2 = 0\\ \Leftrightarrow 2\sin x\cos x + 1 - 2{\sin ^2}x + 3\sin x - \cos x - 2 = 0\\ \Leftrightarrow \cos x\left( {2\sin x - 1} \right) - 2{\sin ^2}x + 3\sin x - 1 = 0\\ \Leftrightarrow \cos x\left( {2\sin x - 1} \right) - \left( {2{{\sin }^2}x - 3\sin x + 1} \right) = 0\\ \Leftrightarrow \cos x\left( {2\sin x - 1} \right) - \left( {2\sin x - 1} \right)\left( {\sin x - 1} \right) = 0\\ \Leftrightarrow \left( {2\sin x - 1} \right)\left( {\cos x - \sin x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = \dfrac{1}{2}\\ \sin x - \cos x = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \sin x = \dfrac{1}{2}\\ \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = \dfrac{1}{2}\\ \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\ x - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x = \dfrac{\pi }{2} + k2\pi \\ x = \pi + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
