Answer
$\left\{300^o+720^o(n), 4210^o+720^o(n), \text{where n is any integer}\right\}$
Work Step by Step
Divide $2\sqrt3$ to both sides of the equation:
\begin{array}{cc}\dfrac{2\sqrt3\cos{(\frac{\theta}{2})}}{2\sqrt3}&=\dfrac{-1}{2\sqrt3}
\\\cos{(\frac{\theta}{2})}&=-\frac{\sqrt3}{2}
\\\frac{\theta}{2}&=\cos^{-1}{(-\frac{\sqrt2}{2})}
\end{array}
Use the inverse cosine function of a scientific calculator to obtain:
\begin{array}{ccc}
&\frac{\theta}{2}=150^o &\text{or} &\frac{\pi}{2}= 210^o
\\&\theta=150^o(2) &\text{ or } &\theta=210^o(2)
\\&\theta=300^o &\text{ or } &\theta=420^o
\end{array}
Since the period of $y=\cos{(\frac{\pi}{2})}$ is $720^o$, then adding a multiple of $720^o$ to each solution of the given equation yields another solution.
Therefore, the complete set of solutions to the given equation is:
$\left\{300^o+720^o(n), 4210^o+720^o(n), \text{where n is any integer}\right\}$
