Answer
$\theta=\{60^o,300^o\}$
Work Step by Step
$sin(\frac{\theta}{2})-cos(\theta)=0$
$sin(\frac{\theta}{2})-[1-2sin^2(\frac{\theta}{2})]=0$
$sin(\frac{\theta}{2})-1+2sin^2(\frac{\theta}{2})=0$
$2sin^2(\frac{\theta}{2})+sin(\frac{\theta}{2})-1=0$
$sin(\frac{\theta}{2})=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(1) \pm \sqrt{(1)^2 - 4 (2)(-1)}}{2(2)}=-1,\frac{1}{2}$
$sin(\frac{\theta}{2})=-1$
$\frac{\theta}{2}=sin^{-1}(1)$
$\frac{\theta}{2}=270^o$
$\theta=540^o+270^on\;\;\;$Reduce
$sin(\frac{\theta}{2})=\frac{1}{2}$
$\frac{\theta}{2}=sin{-1}(\frac{1}{2})$
We know $sin(\frac{\theta}{2})$ is positive in quadrant $I$ and $II$
$\frac{\theta}{2}=30^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-30^o=150^o$
$\theta=60^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=300^o$ Reduce
$\theta=\{60^o,300^o\}$
