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