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