$cos\Bigg(2x+\dfrac{\pi}{6}\Bigg)+cosx=0$

$⇔ 2cos\Bigg(\dfrac{2x+\dfrac{\pi}{6}+x}{2}\Bigg).cos\Bigg(\dfrac{2x+\dfrac{\pi}{6}-x}{2}\Bigg)=0$

$⇔ \left[ \begin{array}{l}cos\Bigg(\dfrac{3x+\dfrac{\pi}{6}}{2}\Bigg)=0\\cos\Bigg(\dfrac{x+\dfrac{\pi}{6}}{2}\Bigg)=0\end{array} \right.$

$⇔ \left[ \begin{array}{l}\dfrac{18x+\pi}{12}=\dfrac{\pi}{2}+k\pi\\\dfrac{6x+\pi}{12}=\dfrac{\pi}{2}+k\pi\end{array} \right.$

$⇔ \left[ \begin{array}{l}x=\dfrac{5\pi}{18}+\dfrac{2k\pi}{3}\\x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.$ $(k∈Z)$

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Công thức biến đổi tổng thành tích: $cosa+cosb=2cos\dfrac{a+b}{2}cos\dfrac{a-b}{2}$

$\begin{array}{l}
\cos \left( {2x + \dfrac{\pi }{6}} \right) + \cos x = 0\\
 \Leftrightarrow 2\cos \left( {\dfrac{{2x + x + \dfrac{\pi }{6}}}{2}} \right)\cos \left( {\dfrac{{2x - x + \dfrac{\pi }{6}}}{2}} \right) = 0\\
 \Leftrightarrow 2\cos \left( {\dfrac{{3x}}{2} + \dfrac{\pi }{{12}}} \right)\cos \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {\dfrac{{3x}}{2} + \dfrac{\pi }{{12}}} \right) = 0\\
\cos \left( {\dfrac{x}{2} + \dfrac{\pi }{{12}}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{3x}}{2} + \dfrac{\pi }{{12}} = \dfrac{\pi }{2} + k\pi \\
\dfrac{x}{2} + \dfrac{\pi }{{12}} = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{3x}}{2} = \dfrac{{5\pi }}{{12}} + k\pi \\
\dfrac{x}{2} = \dfrac{{5\pi }}{{12}} + k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{{5\pi }}{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{18}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$