Answer
{18.43°,135,198.43°,315°}
Work Step by Step
$csc^2~\theta - 2~cot~\theta = 4$
$1+cot^2\theta-2cot \theta=4$ use the identity $1+cot^3\theta=csc^2\theta$
$1+cot^2\theta-2cot \theta-4=0$
$cot^2\theta-2cot \theta-3=0$
$u^2-2u-3=0$ let $cot=u$
solve the equation
u=3=$cot\theta$ I and III Q
u=-1=$cot\theta$ II and IV Q
$arcot(3)=18.43$ and $18.43+180$
$arcot(-1)=135$ and $135+180$
