sin2x=2sinxcosxcos2x=2cos²x-1所以f(x)=5/2sin2x-5√3(1+cos2x)/2+5√3/2=(5/2)sin2x-(5√3/2)cos2x=5(1/2sin2x-√3/2cos2x)=5(sin2xcosπ/6-cos2xsinπ/6)=5sin(2x-π/6)1<=sin(2x-π/6)<=1所以值域[-5,5]