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Phương trình tương đương với:
$2\cos^2(\dfrac{\pi}{4}-2x)-1+\sqrt3\cos4x=2(2\cos^2x-1)$
$\Leftrightarrow \cos(\dfrac{\pi}{2}-4x)+\sqrt3\cos4x=2\cos2x$
$\Leftrightarrow \sin4x+\sqrt3\cos4x=2\cos2x$
$\Leftrightarrow \dfrac{1}{2}\sin4x+\dfrac{\sqrt3}{2}\cos4x=\cos2x$
$\Leftrightarrow \cos(4x-\dfrac{\pi}{6})=\cos2x$
$\Leftrightarrow \left[\begin{array}{l}4x-\dfrac{\pi}{6}=2x+k2\pi\\4x-\dfrac{\pi}{6}=-2x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{36}+k\dfrac{\pi}{3}\end{array}\right.,k\in\mathbb{Z}$
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