f(x)=2sinx(x+π/3)-2sinx
=2[sinxcos(π/3)+cosxsin(π/3)]-2sin2x
=sinx+√3cosx-2sinx
=√3cosx-sinx
=2cos(x+π/6)
x∈[-π/2,0]
x+π/6∈[-π/3,π/6]
所以f(x)的最大值是2,最小值是1
所以值域是[1,2]
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