Let $A\equiv \left(-2,0\right), B=\left(-1,\frac{1}{\sqrt{3}}\right), C\equiv \left(\cos \theta ,\sin \theta \right)$

Equation to $AB,$

$y-0=\left(\frac{{\displaystyle \frac{1}{\sqrt{3}}}-0}{-1+2}\right)\left(x+2\right)$

$\Rightarrow \sqrt{3}y=x+2,$ Point C lies on this line.

$\Rightarrow \sqrt{3}\sin \theta -\cos \theta =2$

$\Rightarrow \frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta =1$

$\Rightarrow \sin \left(\theta -\frac{\pi }{6}\right)=1,$ where $\theta \in \left[0,2\pi \right]$

$\theta -\frac{\pi }{6}=\frac{\pi }{2}$

$\Rightarrow \theta =\frac{\pi }{2}+\frac{\pi }{6}=\frac{2\pi }{3}$

So, only one value of $\theta$ is possible.