Answer
$\theta=\{51.8^o,308.2^o\}$
Work Step by Step
$sin^2(\theta)-cos(\theta)=0$
$1-cos^2(\theta)-cos(\theta)=0$
$cos^2(\theta)+cos(\theta)-1=0$
$cos(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(1) \pm \sqrt{(1)^2 - 4 (1)(-1)}}{2(1)}=\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$
$cos(\theta)=\frac{-1+\sqrt{5}}{2}=0.61$
$\theta=cos^{-1}(\frac{-1+\sqrt{5}}{2})$
We know $cos(\theta)$ is positive in quadrant $I$ and $IV$
$\theta=51.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-51.8^o$
$\theta=51.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=308.2^o$
$cos(\theta)=\frac{-1-\sqrt{5}}{2}=1.6\;\;\;\;$
$\theta=\{51.8^o,308.2^o\}$
