（1）f(x)=√3/2×sin2x-cos²x-1/2=√3/2×sin2x-(2cos²x-1)/2-1=√3/2×sin2x-cos2x/2-1=sin2x×cosπ/6-cos2x×sinπ/6 -1=sin(2x-π/6) -1-1≤sin(2x-π/6)≤1,最小正周期＝π所以,f(x)的最小值＝-1-1...