The solution set of the equation $4\sin \theta \cos \theta -2\cos \theta -2\sqrt{3}\sin \theta +\sqrt{3}=0$ in the interval $(0,2\pi ),$ is

We have

$\begin{array}{l}4\sin \theta \cos \theta -2\cos \theta -2\sqrt{3}\sin \theta +\sqrt{3}=0\\ \Rightarrow 2\sin \theta (2\cos \theta -\sqrt{3})-1(2\cos \theta -\sqrt{3})=0\\ \Rightarrow (2\cos \theta -\sqrt{3})(2\sin \theta -1)=0\\ \Rightarrow \cos \theta =\frac{\sqrt{3}}{2}\text{ or, }\sin \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}\end{array}$