Answer
$\frac{2\pi}{3},\frac{4\pi}{3}$
Work Step by Step
1. $cos(2\theta)+5cos(\theta)+3=0 \Longrightarrow 2cos^2(\theta)-1+5cos(\theta)+3=0\Longrightarrow 2cos^2\theta+5cos\theta+2=0\Longrightarrow (2cos\theta+1)(cos\theta+2)=0 \Longrightarrow cos\theta=-2, -\frac{1}{2}$
2. For $cos\theta=-2$, no real solution.
3. For $cos\theta=-\frac{1}{2}$, we have $\theta=2k\pi+\frac{2\pi}{3}$ or $2k\pi+\frac{4\pi}{3}$,
4. Within $[0,2\pi)$, we have $\theta=\frac{2\pi}{3},\frac{4\pi}{3}$
