Answer
$\theta=\{120^o,240^o\}$
Work Step by Step
$13cot(\theta)+11sec(\theta)=6sin(\theta)$
$13\frac{cos(\theta)}{sin(\theta)}+11\frac{1}{cos(\theta)}=6sin(\theta)$
$13cos(\theta)+11=6sin^2(\theta)$
$13cos(\theta)+11=6[1-cos^2(\theta)]$
$13cos(\theta)+11=6-6cos^2(\theta)$
$6cos^2(\theta)+13cos(\theta)+5=0$
$cos(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(13) \pm \sqrt{(13)^2 - 4 (6)(5)}}{2(6)}=\frac{-5}{3},\frac{-1}{2}$
$cos(\theta)=\frac{-5}{3}$
$\theta=cos^{-1}(\frac{-5}{3})$
$\theta=\varnothing $
$cos(\theta)=\frac{-1}{2}$
$\theta=cos{-1}(\frac{-1}{2})$
We know $cos(\theta)$ is negative in quadrant $II$ and $III$
$\theta=180^o+60^o=240^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-60^o=120^o$
$\theta=240^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=120^o$
$\theta=\{120^o,240^o\}$