Решите уравнение: sin 3x = 1.
\[x =\frac{π}{6} + \frac{2πk}{3}, k∈Z.\]
\[x =±\frac{3π}{4} + 2πk, k∈Z.\]
\[x =\frac{π}{4} + πk, k∈Z.\]
\[x =πk, k∈Z; x = - \frac{π}{4} + πn, n∈Z.\]
\[x = \frac{π}{6} + 2πk, k∈Z; x = \frac{5π}{6} + 2πn, n∈Z.\]