Solve 2cos(θ)+1=02\cos(\theta) + 1 = 02cos(θ)+1=0 for θ∈[0,2π)\theta \in [0, 2\pi)θ∈[0,2π).
θ=π6,11π6\theta = \frac{\pi}{6}, \frac{11\pi}{6}θ=6π,611π
θ=2π3,4π3\theta = \frac{2\pi}{3}, \frac{4\pi}{3}θ=32π,34π
θ=π2,3π2\theta = \frac{\pi}{2}, \frac{3\pi}{2}θ=2π,23π
θ=π3,5π3\theta = \frac{\pi}{3}, \frac{5\pi}{3}θ=3π,35π