$\begin{array}{l} 4\sin x + 4\sin x\cos 2x + 2\sin 2x - 6\cos x - 3 = 0\\ \Leftrightarrow 4\sin x\left( {1 + \cos 2x} \right) + 4\sin x\cos x - 6\cos x - 3 = 0\\ \Leftrightarrow 4\sin x\left( {2{{\cos }^2}x - 1 + 1} \right) + 4\sin x\cos x - 6\cos x - 3 = 0\\ \Leftrightarrow 8\sin x{\cos ^2}x + 4\sin x\cos x - 6\cos x - 3 = 0\\ \Leftrightarrow 4\sin x\cos x\left( {2\cos x + 1} \right) - 3\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left( {2\cos x + 1} \right)\left( {4\sin x\cos x - 3} \right) = 0\\ \Leftrightarrow \left( {2\cos x + 1} \right)\left( {2\sin 2x - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = - \dfrac{1}{2}\\ \sin 2x = \dfrac{3}{2}(\text{Loại}) \end{array} \right.\\ \Rightarrow \cos x = - \dfrac{1}{2} \Leftrightarrow x = \pm \dfrac{{2\pi }}{3} + k2\pi \left( {k \in \mathbb{Z}} \right) \end{array}$
