Solve 2sin2(θ)−sin(θ)−1=02\sin^2(\theta) - \sin(\theta) - 1 = 02sin2(θ)−sin(θ)−1=0 for θ∈[0,2π)\theta \in [0, 2\pi)θ∈[0,2π).
θ=π2,7π6,11π6\theta = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}θ=2π,67π,611π
θ=π2\theta = \frac{\pi}{2}θ=2π
θ=3π2,π6,5π6\theta = \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}θ=23π,6π,65π
θ=π6,5π6\theta = \frac{\pi}{6}, \frac{5\pi}{6}θ=6π,65π