Answer
$\theta=\{218.2^o,321.8^o\}$
Work Step by Step
$cos^2(\theta)+sin(\theta)=0$
$1-sin^2(\theta)+sin(\theta)=0$
$sin^2(\theta)-sin(\theta)-1=0$
$sin(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-1) \pm \sqrt{(-1)^2 - 4 (1)(-1)}}{2(1)}=\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}$
$sin(\theta)=\frac{1+\sqrt{5}}{2}=-0.61$
$\theta=sin^{-1}(\frac{1+\sqrt{5}}{2})$
We know $sin(\theta)$ is negative in quadrant $III$ and $IV$
$\theta=180^o+38.2^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-38.2^o$
$\theta=218.2^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=321.8^o$
$sin(\theta)=\frac{1-\sqrt{5}}{2}=1.6\;\;\;\;$
$\theta=\{218.2^o,321.8^o\}$
