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Nhận xét $ (\frac{\pi}{6}-x )+(x+\frac{\pi}{3} )=\frac{\pi}{2}\implies \cos (\frac{\pi}{6}-x )=\sin(x+\frac{\pi}{3} )$
PT $\iff 1-2\sin^2(x+\frac{\pi}{3} )+4\sin(x+\frac{\pi}{3} )=0\Leftrightarrow \left[ {\begin{matrix} \sin(x+\frac{\pi}{3} )=1-\sqrt{\frac{3}{2}}\\ \sin(x+\frac{\pi}{3} )=1+\sqrt{\frac{3}{2}} (\text{loại}) \end{matrix}} \right.$
$\iff \left[ {\begin{matrix} x=\arcsin\left (1-\sqrt{\frac{3}{2}} \right )-\frac{\pi}{3}+k2\pi\\ x=-\arcsin\left (1-\sqrt{\frac{3}{2}} \right )+\frac{2\pi}{3}+k2\pi \end{matrix}}
\right.$
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