Đáp án:
$\left[ \begin{array}{l}
x = \frac{\pi }{8} + \frac{{k\pi }}{4}\\
x = \frac{\pi }{3} + k\pi \\
x = - \frac{\pi }{3} + k\pi
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
{\sin ^2}x + {\sin ^2}2x + {\sin ^2}3x = \frac{3}{2}\\
\Leftrightarrow \frac{{1 - \cos 2x}}{2} + \frac{{1 - \cos 4x}}{2} + \frac{{1 - \cos 6x}}{2} = \frac{3}{2}\\
\Leftrightarrow \frac{3}{2} - \frac{1}{2}\left( {\cos 2x + \cos 4x + \cos 6x} \right) = \frac{3}{2}\\
\Leftrightarrow \cos 2x + \cos 4x + \cos 6x = 0\\
\Leftrightarrow \left( {\cos 2x + \cos 6x} \right) + \cos 4x = 0\\
\Leftrightarrow 2\cos 4x\cos 2x + \cos 4x = 0\\
\Leftrightarrow \cos 4x\left( {2\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 4x = 0\\
\cos 2x = - \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = \frac{\pi }{2} + k\pi \\
2x = \frac{{2\pi }}{3} + k2\pi \\
2x = - \frac{{2\pi }}{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{8} + \frac{{k\pi }}{4}\\
x = \frac{\pi }{3} + k\pi \\
x = - \frac{\pi }{3} + k\pi
\end{array} \right.
\end{array}$
