f(x)=2sin^x+根号3sin2x-1
=-(1-2sin^2x)+根号3sin2x
=-cos2x+根号3sin2x
=2sin(2x-π/6)
所以
f(x-π/6)=2sin(2x-π/3-π/6)
=2sin(2x-π/2)
=-2cos2x
所以f(x-π/6)+4sinx+1
=-2cos2x+4sinx+1
=-2(1-sin^2x)+4sinx+1
=2sin^2x+4sinx-1
=2(sin^2x+2sinx+1-1)-1
=2(sinx+1)^2 -3=a
又因为x∈[π/6,π/2]
(sinx+1)^2=1/2(a+3)
sinx∈[1/2,1]
sinx+1∈[3/2,2]
(sinx+1)^2∈[9/4,4]
所以a∈[9/4,4]