Answer
$\frac{\pi}{3},\frac{5\pi}{3}$
Work Step by Step
Step 1. Rewrite the equation $sec^2-1=\frac{3}{2}sec\theta \longrightarrow 2sec^2\theta-3sec\theta-2=0 \longrightarrow (2sec\theta+1)(sec\theta-2)=0$. Thus $sec\theta=-\frac{1}{2}$ or $sec\theta=2$.
Step 2. For $sec\theta=\frac{1}{\cos\theta}=2$, we have $cos\theta=\frac{1}{2}$ and $\theta=\frac{\pi}{3},\frac{5\pi}{3}$
Step 3. For $sec\theta=\frac{1}{\cos\theta}=-\frac{1}{2}$, we have $\theta=none$
