equation interval [0,2π)

2cos^2 theta + sin theta-2=0

$$2(1-sin^2\theta) +sin\theta-2=0\\\\
2-2sin^2\theta +sin\theta-2=0\\\\
-2sin^2\theta +sin\theta=0\\\\
sin\theta(-2sin\theta+1)=0\\\\
sin\theta=0\qquad or\qquad -2sin\theta+1=0\\\\
\theta=0,\pi,2\pi,...\;\; or\;\;sin\theta=1/2\\\\
\theta=0,\pi,2\pi,...\;\; or\;\;\pi/6,\;\; 5\pi/6\\\\
so\; for\; the \;domain\; [0,2\pi)\\\\
\theta=0,\;\frac{\pi}{6}, \;\frac{5\pi}{6}\;and\;\pi\\\\$$

you beat me Chris :(

2cos^2 theta + sin theta-2=0

We can write cos^2(θ) as  1 - sin^2(θ)...so we have

2(1 - sin^2(θ)) + sin(θ) - 2 = 0

-2sin^2(θ) + sin(θ) = 0  ... multiply through by -1 and we have

2sin^2(θ) - sin(θ) = 0    factoring, we have...

sin(θ)[2sin(θ) - 1] = 0

Setting the first factor to 0, we have that sin(θ) = 0 ....... and this is true when θ = 0, pi

And setting the second factor to 0, we have 2sin(θ) - 1 = 0

Add 1 to both sides

2sin(θ) = 1   Divide by 2 on both sides

sin(θ) = 1/2   .... and this is true when θ = pi/6 and θ = 5pi/6

So our answers are  θ = [ 0, pi/6, 5pi/6, pi ] on the interval [0, 2pi)