[tex]\bf 1+cos(\theta )=\cfrac{\sqrt{3}+2}{2}\implies cos(\theta )=\cfrac{\sqrt{3}+2}{2}-1
\\\\\\
cos(\theta )=\cfrac{\sqrt{3}+2}{2}-\cfrac{2}{2}\implies cos(\theta )=\cfrac{\sqrt{3}\underline{+2-2}}{2}\implies cos(\theta )=\pm\cfrac{\sqrt{3}}{2}
\\\\\\
cos^{-1}[cos(\theta )]=cos^{-1}\left( \pm\cfrac{\sqrt{3}}{2} \right)\implies \measuredangle \theta =
\begin{cases}
\frac{\pi }{6}\\\\
\frac{5\pi }{6}\\\\
\frac{7\pi }{6}\\\\
\frac{11\pi }{6}
\end{cases}[/tex]
