2nπ2 n \pi2nπ
(2n+1)π3(2 n+1) \frac{\pi}{3}(2n+1)3π
2nπ3\frac{2 n \pi}{3}32nπ
(2n−1)π3(2 n-1) \frac{\pi}{3}(2n−1)3π
No correct answer