Answer
$\theta=\{22.5^o,112.5^o,202.5^o,292.5^o\}$
Work Step by Step
$cot(2\theta)=1$
$\frac{1}{tan(2\theta)}=1$
$tan(2\theta)=1$
$2\theta=tan^{-1}(1)$
We know $tan(\theta)$ is positive in quardent $I$ and quardent $III$
The period of the tangent function is $360^o$
$2\theta=45^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o+45^o\;\;\;\;or\;\;\;\;\;\;2\theta=180^o+45^o\;\;\;\;or\;\;\;\;\;\;2\theta=360^o+135^o$
$\theta=22.5^o\;\;\;\;or\;\;\;\;\;\;\theta=202.5^o\;\;\;\;or\;\;\;\;\;\;\theta=112.5^o\;\;\;\;or\;\;\;\;\;\;\theta=292.5^o\;\;\;$
$\theta=\{22.5^o,112.5^o,202.5^o,292.5^o\}$