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Ta có
$\sin (\frac{\pi}{4}+\frac{3x}{2})=\sin \left[ {\pi -3(\frac{\pi}{4}-\frac{x}{2})} \right]=\sin \left[ {3(\frac{\pi}{4}-\frac{x}{2})} \right]=3\sin (\frac{\pi}{4}-\frac{x}{2})-4\sin^3(\frac{\pi}{4}-\frac{x}{2})$
Vậy PT
$\Leftrightarrow 3\sin (\frac{\pi}{4}-\frac{x}{2})-4\sin^3(\frac{\pi}{4}-\frac{x}{2})=3\sin (\frac{\pi}{4}-\frac{x}{2})$
$\Leftrightarrow 4\sin^3(\frac{\pi}{4}-\frac{x}{2})=0$
$\Leftrightarrow \sin(\frac{\pi}{4}-\frac{x}{2})=0$
$\Leftrightarrow x=\frac{\pi}{2}-k2\pi (k \in \mathbb{Z}).$
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