Solve sin(θ)=32\sin(\theta) = \frac{\sqrt{3}}{2}sin(θ)=23 for θ∈[0,2π)\theta \in [0, 2\pi)θ∈[0,2π).
θ=π3,2π3\theta = \frac{\pi}{3}, \frac{2\pi}{3}θ=3π,32π
θ=π6,5π6\theta = \frac{\pi}{6}, \frac{5\pi}{6}θ=6π,65π
θ=π3,4π3\theta = \frac{\pi}{3}, \frac{4\pi}{3}θ=3π,34π
θ=π6,7π6\theta = \frac{\pi}{6}, \frac{7\pi}{6}θ=6π,67π