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Chú ý rằng
$\cos (\frac{\pi}{6}-x )=\sin \left (\frac{\pi}{2}-(\frac{\pi}{6}-x) \right )=\sin (x+\frac{\pi}{3} )$
$\cos 2(x+\frac{\pi}{3} )=1-2\sin^2 (x+\frac{\pi}{3} )$.
Như vậy
PT $\Leftrightarrow 1-2\sin^2 (x+\frac{\pi}{3} )+4\sin (x+\frac{\pi}{3} )=\frac{5}{2}$
$\Leftrightarrow \sin^2 (x+\frac{\pi}{3} )-2\sin (x+\frac{\pi}{3} )+\frac{3}{4}=0$
$\Leftrightarrow \sin (x+\frac{\pi}{3} )=\frac{1}{2}$
$\Leftrightarrow\left[ {\begin{matrix} x=-\frac{\pi}{6}+k2\pi\\ x=\frac{\pi}{2}+k2\pi \end{matrix}} \right. (k\in \mathbb{Z}).$
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