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