Solve for θ\thetaθ in [0,2π)[0, 2\pi)[0,2π) given the equation 2sin2(θ)−3sin(θ)+1=02\sin^2(\theta) - 3\sin(\theta) + 1 = 02sin2(θ)−3sin(θ)+1=0.
π6,5π6,π2\frac{\pi}{6}, \frac{5\pi}{6}, \frac{\pi}{2}6π,65π,2π
π3,2π3\frac{\pi}{3}, \frac{2\pi}{3}3π,32π
π6,5π6\frac{\pi}{6}, \frac{5\pi}{6}6π,65π
π2\frac{\pi}{2}2π