Answer
$\frac{2\pi}{3}, \frac{4\pi}{3}$
Work Step by Step
Step 1. Rewrite the equation to get $2cos^2\theta-1+5cos\theta+3=0 \longrightarrow 2cos^2\theta+5cos\theta+2=0 \longrightarrow (2cos\theta+1)(cos\theta+2)=0 \longrightarrow cos\theta=-\frac{1}{2}, -2$
Step 2. For $cos\theta=-\frac{1}{2}$, we have $\theta=2k\pi+\frac{2\pi}{3}$ and $\theta=2k\pi+\frac{4\pi}{3}$
Step 3. For $cos\theta=-2$, there is no real solution.
Step 4. Within $[0,2\pi)$, we have $\theta=\frac{2\pi}{3}, \frac{4\pi}{3}$
