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