Giải phương trình sau : $2 \cos \left(2 x+\frac{\pi}{6}\right)+1=0$

$2 \cos \left(2 x+\frac{\pi}{6}\right)+1=0$

$\Leftrightarrow \cos \left(2 x+\frac{\pi}{6}\right)=\frac{-1}{2}$

$\Leftrightarrow \cos \left(2 x+\frac{\pi}{6}\right)=\cos \frac{2 \pi}{3}$

$\Leftrightarrow\left[\begin{array}{l}2 x+\frac{\pi}{6}=\frac{2 \pi}{3}+k 2 \pi \\ 2 x+\frac{\pi}{6}=-\frac{2 \pi}{3}+k 2 \pi\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{l}2 x=\frac{\pi}{2}+k 2 \pi \\ 2 x=\frac{-5 \pi}{6}+k 2 \pi\end{array}\right.$

$\Leftrightarrow\left[\begin{array}{l}x=\frac{\pi}{4}+k \pi \\ x=\frac{-5 \pi}{12}+k \pi\end{array} \quad(k \in \mathbb{Z})\right.$