f(x)=2sinxcosx-根号3cos2x+1 =sin2x-√3cos2x+1=2(sin2x×(1/2)-cos2x×(√3/2))+1=2sin(2x-π/3)+1所以最小正周期T=2π/2=π;您好,很高兴为您解答,skyhunter002为您答疑解惑如果本题有什么不明白可以追问,如果满意...