$cos(x+\dfrac{\pi}{3}=-\dfrac{1}{2}$
$⇔cos(x+\dfrac{\pi}{3}=cos(\dfrac{2\pi}{3})$
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=-\pi+k2\pi\end{array} \right.\)
Thay $x=\dfrac{\pi}{3}+k2\pi$ vào $(0;6\pi)$
$0<\dfrac{\pi}{3}+k2\pi<6\pi$
$⇔-\dfrac{\pi}{3}<k2\pi<\dfrac{17\pi}{3}$
$⇔-\dfrac{1}{6}<k<\dfrac{17}{6}$
$k∈Z⇒k=\{0;1;2\}$
Thay $x=-\pi+k2\pi$ vào $(0;6\pi)$
$0<-\pi+k2\pi<6\pi$
$⇔\pi<k2\pi<7\pi$
$⇔\dfrac{1}{2}<k<\dfrac{7}{2}$
$⇔k=\{0;1;2;3\}$
Vậy ta chọn $D.7$