Answer
$\theta=\{7.5^o+45^ok\;\;,\;\;37.5^o+45^ok\}$
Work Step by Step
$cos(8\theta)=\frac{1}{2}$
$8\theta=cos^{-1}(\frac{1}{2})$
We know $cos(\theta)$ is Positive in quardent $I$ and quardent $IV$
The period of the cosine function is $360^o$
$8\theta=60^o\;\;\;\;\;\;\;or\;\;\;\;\;\;\;8\theta=360^o-60^o$
$8\theta=60^o+360^ok\;\;\;\;\;\;\;or\;\;\;\;\;\;\;8\theta=300^o+360^ok $
$\theta=7.5^o+45^ok\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\theta=37.5^o+45^ok $
$\theta=\{7.5^o+45^ok\;\;,\;\;37.5^o+45^ok\},\;\;\;\;\;\;\;\;\;\;when \;\;\;K=\{0,1,2,....\}$
