Question

已知函数y=f(x)的定义域是【0,1/4】,求函数f(sin^2x)的定义域？

Answer 1

由条件知sin²x≤1/4,-1/2≤sinx≤1/2
由图像知,-π/6+2kπ≤x≤π/6+2kπ
所以f(sin²x)的定义域为[-π/6+2kπ,π/6+2kπ],k∈Z,10,0<=sin^2x<=1/4
-1/2<=sinx<=1/2
kπ-π/6<=x<=kπ+π/6,0,