f(x)=根号3sin^2+sinxcosx
=根号3(1-sin2x/2)+sin2x/2
=根号3/2-根号3/2cos2x+1/2sin2x
=根号3/2+sin(2x-π/3)
因为x∈[π/2,π],
所以2x∈[π,2π],
2x-π/3∈[2π/3,5π/3],
sin(2x-π/3)∈[-1,根号3/2]
所以f(x)∈[根号3/2-1,根号3]
f(x)=根号3sin^2+sinxcosx
=根号3(1-sin2x/2)+sin2x/2
=根号3/2-根号3/2cos2x+1/2sin2x
=根号3/2+sin(2x-π/3)
因为x∈[π/2,π],
所以2x∈[π,2π],
2x-π/3∈[2π/3,5π/3],
sin(2x-π/3)∈[-1,根号3/2]
所以f(x)∈[根号3/2-1,根号3]