Precalculus (6th Edition)

$\{45^{o},225^{o}\}$
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= $\{45^{o},225^{o}\}$