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$\sin (\frac{3\pi}{5}+2x )=2\sin (\frac{\pi}{5}-x )$
$\Leftrightarrow \sin(\frac{2\pi}{5}-2x)=2\sin(\frac{\pi}{5}-x)$
$\Leftrightarrow 2\sin(\frac{\pi}{5}-x)
\cos(\frac{\pi}{5}-x)=2\sin(\frac{\pi}{5}-x) $
$\Leftrightarrow \left[ \begin{array}{l}\sin(\frac{\pi}{5}-x)=0\\
\cos(\frac{\pi}{5}-x)=1 \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} \frac{\pi}{5}-x=k\pi\\ \frac{\pi}{5}-x=2k\pi \end{array} \right. (k\in\mathbb{Z})$
$\Leftrightarrow x=\frac{\pi}{5}-k\pi,k\in\mathbb{Z}.$
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