急!设函数f(x)=√3sin2x-sin^2x-3cos^2x

f(x)=√3sin2x-sin^2x-3cos^2x
=√3sin2x-1-2cos²x
=√3sin2x-(cos2x+1)-1
=2sin(2x-π/6)-2
当 x=0时 2x-π/6=-π/6
所以初相为 -π/6
x∈【0,π/2】
2x-π/6∈[-π/6,5π/6]
所以
sin(2x-π/6)∈[-1/2,1]
则 f(x)的值域为 [-3,0]