Solve sin(2θ)=cos(θ)\sin(2\theta) = \cos(\theta)sin(2θ)=cos(θ) for θ∈[0,2π)\theta \in [0, 2\pi)θ∈[0,2π).
θ=π2,3π2\theta = \frac{\pi}{2}, \frac{3\pi}{2}θ=2π,23π
θ=0,π2,π\theta = 0, \frac{\pi}{2}, \piθ=0,2π,π
θ=π3,2π3\theta = \frac{\pi}{3}, \frac{2\pi}{3}θ=3π,32π
θ=π6,5π6,3π2\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}θ=6π,65π,23π