Sin^2x+sin^2(3x)=2sin^2(2x)

Đáp án:

$\begin{array}{l}
{\sin ^2}x + {\sin ^2}3x = 2{\sin ^2}2x\\
\Leftrightarrow {\sin ^2}x – {\sin ^2}2x = {\sin ^2}2x – {\sin ^2}3x\\
\Leftrightarrow (\sin x – \sin 2x)(\sin x + \sin 2x) = (\sin 2x – \sin 3x)(\sin 2x + \sin 3x)\\
\Leftrightarrow 2.\cos \frac{{3x}}{2}.\sin \frac{{ – x}}{2}.2.\sin \frac{{3x}}{2}.\cos \frac{x}{2} = 2.\cos \frac{{5x}}{2}.\sin \frac{{ – x}}{2}.2.\sin \frac{{5x}}{2}.\cos \frac{{ – x}}{2}\\
\Leftrightarrow \sin 3x.( – \sin x) = \sin 5x.( – \sin x)\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin 3x = \sin 5x
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
3x = 5x + k2\pi \\
3x = \pi – 5x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{8} + \frac{{k\pi }}{4}
\end{array} \right.(k \in Z)
\end{array}$