`~rai~`
\(cos\left(2x+\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\\\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=cos\left(\dfrac{2\pi}{3}\right)\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\\2x+\dfrac{\pi}{3}=-\dfrac{2\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\pi+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy phương trình có các họ nghiệm là:}x=\dfrac{\pi}{6}+k\pi;x=-\dfrac{\pi}{2}+k\pi.(k\in\mathbb{Z})\)
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