已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,当x∈[π/12,7π/12]时求f-1（1）的值
解析：∵函数f(x)=2cosxsin(x+π/3)-√3sin^2x+sinxcosx
=2cosx(1/2sinx+√3/2cosx)-√3/2(1-cos2x)+1/2sin2x
=1/2sin2x+√3/2(1+cos2x)-√3/2(1-cos2x)+1/2sin2x
=sin2x+√3cos2x=2sin(2x+π/3)
∴f(x)=2sin(2x+π/3),其定义域为[π/12,7π/12]
∴值域为[-2,2]
其反函数为f^(-1)(x)=[ π-arcsin(x/2)- π/3]/2,其定义域为[-2,2],值域为[π/12,7π/12]
f^(-1)(1)=[ π-arcsin(1/2)- π/3]/2=π/4