Answer
$\theta=\{36.9^o,48.2^o,311.8^o,323.1^o\}$
Work Step by Step
$23csc^2(\theta)-22cot(\theta)csc(\theta)-15=0$
$23\frac{1}{sin^2(\theta)}-22\frac{cos(\theta)}{sin(\theta)}.\frac{1}{sin(\theta)}-15=0$
$23-22cos(\theta)-15sin^2(\theta)=0$
$23-22cos(\theta)-15[1-cos^2(\theta)]=0$
$23-22cos(\theta)-15+15cos^2(\theta)=0$
$15cos^2(\theta)-22cos(\theta)+8=0$
$cos(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-22) \pm \sqrt{(-22)^2 - 4 (15)(8)}}{2(15)}=\frac{4}{5},\frac{2}{3}$
$cos(\theta)=\frac{4}{5}$
$\theta=cos^{-1}(\frac{4}{5})$
We know $cos(\theta)$ is positive in quadrant $I$ and $IV$
$\theta=36.9^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-36.9^o=323.1^o$
$\theta=36.9^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=323.1^o$
$cos(\theta)=\frac{2}{3}$
$\theta=cos^{-1}(\frac{2}{3})$
We know $cos(\theta)$ is positive in quadrant $I$ and $IV$
$\theta=48.2^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-48.2^o=311.8^o$
$\theta=48.2^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=311.8^o$
$\theta=\{36.9^o,48.2^o,311.8^o,323.1^o\}$
