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$\begin{array}{l}
\left( 1 \right) \Leftrightarrow - {\log _6}\left( {\sin \frac{x}{2} - 3tanx - \frac{{3\sqrt 3 }}{2}}
\right) + {\log _6}\left( {\sin \frac{x}{2} - tan2x - \frac{{3\sqrt 3 }}{2}} \right) = 0\\
\,\,\,\ \Leftrightarrow \sin \frac{x}{2} - 3tanx - \frac{{3\sqrt 3 }}{2} = \sin \frac{x}{2} - tan2x
- \frac{{3\sqrt 3 }}{2} > 0\\
\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
\sin \frac{x}{2} - 3tanx - \frac{{3\sqrt 3 }}{2} > 0 & \left( 2 \right)\\
3tanx = tan2x & & \left( 3 \right)
\end{array} \right.
\end{array}$
Với $t = \tan x$, $\left( 3 \right) \Leftrightarrow 3t = \frac{{2t}}{{1 - {t^2}}}$.
$\Leftrightarrow (3t^2-1)=0\Leftrightarrow \left[ \begin{array}{l}t=0\\t=\pm \frac{1}{\sqrt{3}}\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}x = k\pi (không thỏa mãn) (4)\\x=\pm \frac{\pi}{6}+k\pi\end{array} \right.$
$t=\frac{1}{\sqrt{3}}$ không thỏa mãn điều kiện $(2)$
$t=\frac{-1}{\sqrt{3}}\Leftrightarrow x=\frac{-\pi}{6}+k\pi$
Xét điều kiện $(2)$ :
$\begin{array}{l}
\,\sin \left( { - \frac{\pi }{{12}} + \frac{{k\pi }}{2}} \right) + \frac{{3\sqrt 3 }}{3} - \frac{{3\sqrt
3 }}{2} > 0\\
\,\,\,\,\,\,\,\, \Leftrightarrow \sin \left( { - \frac{\pi }{{12}} + \frac{{k\pi }}{2}} \right) >
\frac{{\sqrt 3 }}{2}
\end{array}$
Suy ra : $k = 4h + 1$, khi đó $x=\frac{5\pi}{6}+4h\pi$ $(h\in Z)$
Phương trình có nghiệm duy nhất:
$x=\frac{5\pi}{6}+4h\pi$ $(h\in Z)$
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