Solve for θ∈[0,2π)\theta \in [0, 2\pi)θ∈[0,2π) if 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
π3,2π3,3π2\frac{\pi}{3}, \frac{2\pi}{3}, \frac{3\pi}{2}3π,32π,23π
π6,5π6,3π2\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}6π,65π,23π
π4,3π4,π2\frac{\pi}{4}, \frac{3\pi}{4}, \frac{\pi}{2}4π,43π,2π