If an=sin(nθ+α)a_n = \sin(n\theta + \alpha)an=sin(nθ+α) is an arithmetic progression, what must be true about θ\thetaθ?
θ=0\theta = 0θ=0
θ=π\theta = \piθ=π
θ=π2\theta = \frac{\pi}{2}θ=2π
θ=2π\theta = 2\piθ=2π