The general solution of \(2 \sqrt{3} \cos ^2 \theta=\sin \theta\) is

\(\begin{aligned} & 2 \sqrt{3} \cos ^2 \theta=\sin \theta \\ & \Rightarrow 2 \sqrt{3}\left(1-\sin ^2 \theta\right)=\sin \theta \\ & \Rightarrow 2 \sqrt{3} \sin ^2 \theta+\sin \theta-2 \sqrt{3}=0 \\ & \Rightarrow(\sqrt{3} \sin \theta+2)(2 \sin \theta-\sqrt{3})=0 \\ & \Rightarrow \sqrt{3} \sin \theta+2=0 \text { or } 2 \sin \theta-\sqrt{3}=0 \\ & \Rightarrow \sin \theta=-\frac{2}{\sqrt{3}}, \text { which is not possible } \\ & \text { or } \sin \theta=\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3} \\ & \Rightarrow \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}}\left(\frac{\pi}{3}\right), \mathrm{n} \in \mathrm{Z}\end{aligned}\)