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$pt \Leftrightarrow \cos6x+\cos4x=\dfrac{\sqrt3}{2}\cos2x+\dfrac{1}{2}\sin2 x+\frac{\sqrt3}{2}\\\Leftrightarrow 2\cos5x\cos x=\cos\left(2x-\dfrac{\pi}{6}\right)+\cos\dfrac{\pi}{6}\\\Leftrightarrow 2\cos5x\cos x=2\cos x\cos \left(x-\dfrac{\pi}{6}\right)\\\Leftrightarrow \left[\begin{array}{1}\cos x=0 \\\cos5x=\cos\left(x-\dfrac{\pi}{2}\right) \end{array}\right.$
Đến đây có thể tự giải rồi.
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