Answer
$\theta=\{0^o,240^o\}$
Work Step by Step
$cos(\frac{\theta}{2})-cos(\theta)=0$
$cos(\frac{\theta}{2})-[2cos^2(\frac{\theta}{2})-1]=0$
$cos(\frac{\theta}{2})-2cos^2(\frac{\theta}{2})+1=0$
$2cos^2(\frac{\theta}{2})-cos(\frac{\theta}{2})-1=0$
$cos(\frac{\theta}{2})=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-1) \pm \sqrt{(-1)^2 - 4 (2)(-1)}}{2(2)}=1,\frac{-1}{2}$
$cos(\frac{\theta}{2})=1$
$\frac{\theta}{2}=cos^{-1}(1)$
$\frac{\theta}{2}=0^o$
$\theta=0^o$
$cos(\frac{\theta}{2})=\frac{-1}{2}$
$\frac{\theta}{2}=cos{-1}(\frac{-1}{2})$
We know $cos(\frac{\theta}{2})$ is negative in quadrant $II$ and $III$
$\frac{\theta}{2}=180^o-60^o=120^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\frac{\theta}{2}=180^o+60^o=240^o$
$\theta=240^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=480^o$
$\theta=\{0^o,240^o\}$
