解∵f(x)=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3(sinx)^2+sinxcosx
=sinxcosx+√3(cosx)^2-√3(sinx)^2+sinxcosx
=sin2x+√3cos2x
=2[(1/2)sin2x+(√3/2)cos2x]
=2[sin2xcos(π/3)+cos2xsin(π/3)]
=2sin[2x+(π/3)]
∴當(dāng)2x+(π/3)=π/2時,函數(shù)f(x)有最大值且f(x)=2,此時,x=π/12；
當(dāng)2x+(π/3)=3π/2時,函數(shù)f(x)有最小值且f(x)=-2,此時,x=7π/12；
又∵sinx的單調(diào)減區(qū)間為：2kπ+（π/2)≤x≤2kπ+（3π/2)；
∴2kπ+（π/2)≤2x+(π/3)≤2kπ+（3π/2)
解之得：kπ+（π/12)≤x≤kπ+（7π/12)