Answer
$41.8^{\circ},48.6^{\circ},131.4^{\circ},138.2^{\circ}$
Work Step by Step
$\dfrac{18}{\cos^2{\theta}} - \dfrac{17 \sin{\theta}}{\cos^2{\theta}} - 12 = 0$
$18 - 17 \sin{\theta} - 12 \cos^2{\theta}=0$
$18-17\sin{\theta} -12 + 12 \sin^2{\theta} =0$
$12 \sin^2{\theta} - 17 \sin{\theta} + 6 = 0$
Using the quadratic formula:
$\sin{\theta} = \dfrac{17 \pm \sqrt{289-4(12)(6)}}{2(12)} = \dfrac{17 \pm 1}{24}$
$\sin{\theta} = \dfrac{18}{24} \hspace{30pt} \sin{\theta} = \dfrac{16}{24}$
$\sin{\theta} = \dfrac{3}{4} \hspace{30pt} \sin{\theta} = \dfrac{2}{3}$
$\theta = 48.6^{\circ},131.4^{\circ} \hspace{30pt} \theta = 41.8^{\circ},138.2 ^{\circ}$
$\theta = 41.8^{\circ},48.6^{\circ},131.4^{\circ},138.2^{\circ}$
