Answer
$\{\frac{\pi}{6},\frac{5\pi}{6} \}$
Work Step by Step
Step 1. Use the identity $cos^2x=1-sin^2x$, we have $4-4sin^2x+4sin(x)-5=0$ or $4sin^2x-4sin(x)+1=0$, or $(2sin(x)-1)^2=0$, which gives solutions as $sin(x)=\frac{1}{2}$ (multiplicity 2)
Step 2. For $sin(x)=\frac{1}{2}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{6},\frac{5\pi}{6}$
Step 3. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{6},\frac{5\pi}{6} \}$
