<s>[INST] Show that $2\cos(\frac{\theta}{2})=\cos \theta$ iff $\theta = (4n+2)\pi \pm 2\phi$ Q:

Show that $2\cos(\frac{\theta}{2})=\cos \theta$ if and only if
$\theta=(4n+2)\pi\pm 2\phi$ where $\phi$ is defined by $\cos(\phi)= \frac{1}{2}(\sqrt 3-1)\;$, $0\le \phi\le \frac{1}{2}\pi$, and $n$
is any integer

Workings
$2\cos(\frac{\theta}{2}) = \cos(\theta) \iff 2\cos(\frac{\theta}{2}) = \cos^2(\frac{\theta}{2}) - \sin^2(\frac{\theta}{2}) \iff 2\cos^2(\frac{\theta}{2})-2\cos(\frac{\theta}{2})-1=0$
So we have, $$ \cos(\frac{\theta}{2}) = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$$
but, $\cos(\frac{\theta}{2}) \not = \frac{1}{2} +\frac{\sqrt{3}}{2}$ > 1
So $\cos(\frac{\theta}{2}) = \frac{1}{2} -\frac{\sqrt{3}}{2}$
$\therefore \frac{\theta}{2} = \pm \cos^{-1}(\frac{1-\sqrt{3}}{2})+2\pi n$
$\therefore \theta = \pm 2\cos^{-1}(\frac{1-\sqrt{3}}{2})+4\pi n$
I notice that the form is similar to what the question wants us to show. But it's not quite the same. Any idea where I've gone wrong?
 [/INST] You have correctly got
$$ \theta = \pm 2\cos^{-1}\bigg(\frac{1-\sqrt{3}}{2}\bigg)+4\pi n$$
Now, note that
$$\cos^{-1}\bigg(\frac{1-\sqrt{3}}{2}\bigg)=\pi-\phi$$
So, one finally has
$$ \theta = \pm 2(\pi-\phi)+4\pi n,$$
i.e.
$$\theta=(4n+2)\pi-2\phi,\ (4n-2)\pi+2\phi$$
where the latter can be written as
$$\theta=(4m+2)\pi+2\phi$$
where $m=n-1$.
Therefore, one can write
$$\theta=(4n+2)\pi\pm 2\phi$$
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