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