Solve cos(2θ)=12\cos(2\theta) = \frac{1}{2}cos(2θ)=21 for θ∈[0,2π)\theta \in [0, 2\pi)θ∈[0,2π).
θ=π12,11π12,13π12,23π12\theta = \frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12}θ=12π,1211π,1213π,1223π
θ=π6,5π6,7π6,11π6\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}θ=6π,65π,67π,611π
θ=π3,5π3\theta = \frac{\pi}{3}, \frac{5\pi}{3}θ=3π,35π
θ=π6,5π6\theta = \frac{\pi}{6}, \frac{5\pi}{6}θ=6π,65π