$\begin{array}{l} \sin 2x + \cos x = \left( {2\sin x + 1} \right)\left( {3\cos x - 1} \right)\\ \Leftrightarrow \sin 2x + \cos x = 6\sin x\cos x - 2\sin x + 3\cos x - 1\\ \Leftrightarrow \sin 2x + \cos x = 3\sin 2x - 2\sin x + 3\cos x - 1\\ \Leftrightarrow 1 - 2\sin 2x = 2\cos x - 2\sin x\\ \Leftrightarrow 2 - 2\sin 2x = 2\left( {\cos x - \sin x} \right) + 1\\ \Leftrightarrow 2\left( {1 - \sin 2x} \right) = 2\left( {\cos x - \sin x} \right) + 1\\ \Leftrightarrow 2\left( {{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x} \right) = 2\left( {\cos x - \sin x} \right) + 1\\ \Leftrightarrow 2{\left( {\cos x - \sin x} \right)^2} = 2\left( {\cos x - \sin x} \right) + 1\\ \Leftrightarrow 2{t^2} - 2t - 1 = 0\left( {t = \cos x - \sin x,\left| t \right| \le \sqrt 2 } \right)\\ \Leftrightarrow \left[ \begin{array}{l} t = \dfrac{{1 + \sqrt 3 }}{2}\\ t = \dfrac{{1 - \sqrt 3 }}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \cos x - \sin x = \dfrac{{1 + \sqrt 3 }}{2}\\ \cos x - \sin x = \dfrac{{1 - \sqrt 3 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{1 + \sqrt 3 }}{2}\\ - \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{1 - \sqrt 3 }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \sin \left( {x - \dfrac{\pi }{4}} \right) = - \dfrac{{\sqrt 6 + \sqrt 2 }}{4}\\ \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 6 - \sqrt 2 }}{4} \end{array} \right. \end{array}$
Phương trình cơ bản, tự tính nốt nhé.
