If f(x) is defined on (0, 1], then the domain of f(sinx) is
2nπ,2n+1π,n∈Z
2n+1π2,2n+3π2; n∈Z
n−1π,n+1π,n∈Z
nπ,2n+1πn∈Z
Since the domain of f is (0, 1]
∴0<sinx≤1⇒2nπ<x<2n+1π,n∈Z