(6n−1)π9(6 n-1) \frac{\pi}{9}(6n−1)9π
2nπ3+π3\frac{2 \mathrm{n} \pi}{3}+\frac{\pi}{3}32nπ+3π
2nπ3±π9\frac{2 n \pi}{3} \pm \frac{\pi}{9}32nπ±9π
(2n+1)π6(2 n+1) \frac{\pi}{6}(2n+1)6π