Answer
$\{45^{o},225^{o}\}$
Work Step by Step
Using the Pythagorean identity $1 +\cot^{2}\theta=\csc^{2}\theta$
the equation becomes
$1 +\cot^{2}\theta-2\cot\theta=0$
... substitute $t=\cot\theta,$
$t^{2}-2t+1=0$
... square of a difference ...
$(t-1)^{2}=0$
$t-1=0$
$t=1$
Back-substitute,
$\cot\theta=1$
$\theta=45^{o},225^{o}$
Solution set= $\{45^{o},225^{o}\}$
