1/4≤x≤1/2
0≤4x-1≤1,所以arccos（4x-1）∈[0,∏/2]
y=√arccos（4x-1）∈[0,根号下∏/2]
y^2=arccos(4x-1)
cos(y^2)=4x-1
x=[cos(y^2)+1]/4
反函数y=[cos(x^2)+1]/4,x∈[0,根号下∏/2]