Answer
$\theta=\{68.5^o,291.5^o\}$
Work Step by Step
$2sin^2(\theta)-2cos(\theta)-1=0$
$2[1-cos^2(\theta)]-2cos(\theta)-1=0$
$2-2cos^2(\theta)-2cos(\theta)-1=0$
$2cos^2(\theta)+2cos(\theta)+1=0$
$cos(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(2) \pm \sqrt{(2)^2 - 4 (2)(1)}}{2(2)}=\frac{-1+\sqrt{3}}{2},\frac{-1-\sqrt{3}}{2}$
$cos(\theta)=\frac{-1-\sqrt{3}}{2}=-1.366\;\;\;\;$
$cos(\theta)=\frac{-1+\sqrt{3}}{2}=0.366$
$\theta=cos^{-1}(\frac{-1+\sqrt{3}}{2})$
We know $cos(\theta)$ is positive in quadrant $I$ and $IV$
$\theta=68.5^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-68.5^o=291.5^o$
$\theta=68.5^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=291.5^o$
$\theta=\{68.5^o,291.5^o\}$
