Solve sin(θ)=−12\sin(\theta) = -\frac{1}{2}sin(θ)=−21 for θ∈[0,2π)\theta \in [0, 2\pi)θ∈[0,2π).
θ=π6,5π6\theta = \frac{\pi}{6}, \frac{5\pi}{6}θ=6π,65π
θ=7π6,11π6\theta = \frac{7\pi}{6}, \frac{11\pi}{6}θ=67π,611π
θ=π3,2π3\theta = \frac{\pi}{3}, \frac{2\pi}{3}θ=3π,32π
θ=2π3,4π3\theta = \frac{2\pi}{3}, \frac{4\pi}{3}θ=32π,34π