θ=(2n+1)π2,n∈Z\theta=(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}θ=(2n+1)2π,n∈Z
θ=(2n+1)π,n∈z\theta=(2 n+1) \pi, n \in zθ=(2n+1)π,n∈z
θ=(4n−1)π2,n∈z\theta=(4 n-1) \frac{\pi}{2}, n \in zθ=(4n−1)2π,n∈z
θ=(4n+1)π2,n∈z\theta=(4 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{z}θ=(4n+1)2π,n∈z
