Answer
$\theta=\{60^o,180^o,300^o\}$
Work Step by Step
$cos(3\theta)=-1$
$3\theta=cos^{-1}(-1)$
We know $cos(\theta)$ is negative in quardent $II$ and quardent $III$
The period of the cosine function is $360^o$
$3\theta=180^o\;\;\;\;or\;\;\;\;\;\;3\theta=360^o+180^o\;\;\;\;or\;\;\;\;\;\;3\theta=720^o+180^o\;\;\;\;\;\;\;\;\;\;$
$\theta=60^o\;\;\;\;or\;\;\;\;\;\;\theta=180\;\;\;\;or\;\;\;\;\;\;\theta=300^o\;\;\;\;\;\;\;\;\;\;\;\;\;$
$\theta=\{60^o,180^o,300^o\}$