You have one value of $\frac 1 2 \theta$, i.e.
\[
\frac 1 2 \theta = \arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right)
\]
However, $\cos x=k$ has infinitely many solutions (draw a sketch!), and if $x=\alpha$ is one
then so is $x=-\alpha $ and $x=\alpha+2\pi$ etc. (look at your sketch!) The general solutions here are $x=\alpha + 2n\pi$ and $x=-\alpha + 2n\pi$. (Look at your sketch to convince yourself!)

Relating this back to your problem, you have the general solutions:
\[
\frac 1 2 \theta = \arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right) + 2n\pi
\]
and
\[
\frac 1 2 \theta = -\arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right) + 2n\pi
\]
This is starting to look a little better - if we multiply by 2 we will get a $4n\pi$ which is promising.

The next bit is to try to relate $\arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right)$ to $\phi$. Have a play and see if you can get this bit out.