f(x)=sinxcosx-√3cos^2x-√3
=1/2sin2x-√3(1+cos2x)/2-√3
=1/2sin2x-√3/2-√3/2cos2x-√3
=1/2sin2x-√3/2cos2x-3√3/2
=sin2xcosπ/3-cos2xsinπ/3-3√3/2
=sin(2x-π/3)-3√3/2
T=2π/2=π
-1<=sin(2x-π/3)<=1
-1-3√3/2<=sin(2x-π/3)-3√3/2<=1-3√3/2
即f(x)∈[-1-3√3/2,1-3√3/2]
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