sin^2(2x)+sin^2(4x)=sin^2(6x)

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