Answer
$\theta=\{30^o,150^o,210^o,330^o\}$
Work Step by Step
$4cos(\theta)-3sec(\theta)=0$
$4cos(\theta)-3\frac{1}{cos(\theta)}=0\;\;\;\;\;\;\;\;\;\;$ multiply each side by $cos(\theta)$.
$4cos^2(\theta)-3=0$
$4cos^2(\Theta )=3\;\;\;\;\;\;\;\;\;\;$ subtract $ -3 $ from each side.
$cos^2(\Theta )=\frac{3}{4} \;\;\;\;\;\;\;\;\;\;\;$ divide each side by $4$
$cos(\Theta )=\pm \sqrt{\frac{3}{4}}$
$cos(\Theta )=\pm \frac{\sqrt{3}}{2}$
$\theta= cos^{-1}(\frac{\sqrt{3}}{2})$
We know $ cos(\theta) $ is positive in quadrant $I$ and quadrant $IV$
$\theta=30^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=360^o-30^o=330^o$
$\theta=30^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=330^o$
$\theta= cos^{-1}(\frac{-\sqrt{3}}{2})$
We know $ cos(\theta) $ is negative in quadrant $II$ and quadrant $III$
$\theta=180^o+30^o=210^o\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\theta=180^o-30^o=150^o$
$\theta=210^o\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\theta=150^o$
$\theta=\{30^o,150^o,210^o,330^o\}$
