Answer
$\theta=\{41.8^o,48.6^o,131.4^o,138.2^o\}$
Work Step by Step
$18sec^2(\theta)-17tan(\theta)sec(\theta)-12=0$
$18\frac{1}{cos^2(\theta)}-17\frac{sin(\theta)}{cos(\theta)}.\frac{1}{cos(\theta)}-12=0$
$18-17sin(\theta)-12cos^2(\theta)=0$
$18-17sin(\theta)-12[1-sin^2(\theta)]=0$
$18-17sin(\theta)-12+12sin^2(\theta)=0$
$12sin^2(\theta)-17sin(\theta)+6=0$
$sin(\theta)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-17) \pm \sqrt{(-17)^2 - 4 (12)(6)}}{2(12)}=\frac{3}{4},\frac{2}{3}$
$sin(\theta)=\frac{3}{4}$
$\theta=sin^{-1}(\frac{3}{4})$
We know $sin(\theta)$ is positive in quadrant $I$ and $II$
$\theta=48.6^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-48.6^o=131.4^o$
$\theta=48.6^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=131.4^o$
$sin(\theta)=\frac{2}{3}$
$\theta=sin{-1}(\frac{2}{3})$
We know $sin(\theta)$ is positive in quadrant $I$ and $II$
$\theta=41.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-41.8^o=138.2^o$
$\theta=41.8^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=138.2^o$
$\theta=\{41.8^o,48.6^o,131.4^o,138.2^o\}$