$\begin{array}{l} i){\sin ^2}x + {\sin ^2}2x + {\sin ^2}3x = \dfrac{3}{2}\\ \Leftrightarrow \dfrac{{1 - \cos 2x}}{2} + \dfrac{{1 - \cos 4x}}{2} + \dfrac{{1 - \cos 6x}}{2} = \dfrac{3}{2}\\ \Leftrightarrow \dfrac{3}{2} - \dfrac{{\left( {\cos 2x + \cos 4x + \cos 6x} \right)}}{2} = \dfrac{3}{2}\\ \Leftrightarrow \cos 2x + \cos 4x + \cos 6x = 0\\ \Leftrightarrow 2\cos 4x\cos 2x + \cos 4x = 0\\ \Leftrightarrow \cos 4x\left( {2\cos 2x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos 4x = 0\\ \cos 2x = - \dfrac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 4x = \dfrac{\pi }{2} + k\pi \\ 2x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\ x = \pm \dfrac{\pi }{3} + k\pi \end{array} \right.\left( {k \in Z} \right) \end{array}$
$\begin{array}{l}
j){\sin ^2}2x + {\sin ^2}4x - {\sin ^2}6x = 0\\
\Leftrightarrow \dfrac{{1 - \cos 4x}}{2} + \dfrac{{1 - \cos 8x}}{2} - \dfrac{{1 - \cos 12x}}{2} = 0\\
\Leftrightarrow \dfrac{1}{2} - \dfrac{{\left( {\cos 4x + \cos 8x - \cos 12x} \right)}}{2} = 0\\
\Leftrightarrow \cos 4x + \cos 8x - \cos 12x - 1 = 0\\
\Leftrightarrow 2\cos 6x\cos 2x - \left( {2{{\cos }^2}6x - 1} \right) - 1 = 0\\
\Leftrightarrow 2\cos 6x\cos 2x - 2{\cos ^2}6x = 0\\
\Leftrightarrow 2\cos 6x\left( {\cos 2x - \cos 6x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 6x = 0\\
\cos 2x = \cos 6x
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
6x = \dfrac{\pi }{2} + k\pi \\
6x = 2x + k2\pi \\
6x = - 2x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{6}\\
x = k\dfrac{\pi }{2}\\
x = \dfrac{{k\pi }}{4}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
$\begin{array}{l}
m)\sin 2x\sin 4x + \cos 6x = 0\\
\Leftrightarrow \dfrac{1}{2}\left[ {\cos 2x - \cos 6x} \right] + \cos 6x = 0\\
\Leftrightarrow \dfrac{1}{2}\cos 2x + \dfrac{1}{2}\cos 6x = 0\\
\Leftrightarrow \cos 2x + \cos 6x = 0\\
\Leftrightarrow 2\cos 4x\cos 2x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 4x = 0\\
\cos 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k\pi \\
2x = \dfrac{\pi }{2} + k\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
