Question

Вопрос №163038

Answer 1

sin2x=1/4;
2x=arcsin(1/4)+2πk, k∈Z    или  2х=π-arcsin(1/4)+2πn, n∈Z;
x=(1/2)·arcsin(1/4)+πk, k∈Z    или  х=(π/2)-(1/2)·arcsin(1/4)+πn, n∈Z.

О т в е т.(1/2)·arcsin(1/4)+πk ; (π/2)-(1/2)·arcsin(1/4)+πn, k, n∈Z.