f(x)=向量m*向量n所以f(x)=（根号3sin2x+2）×1+cosx×2cosx =根号3sin2x+2+1+cos2x =2sin(2x+π\6)+3所以最小正周期为π单调递减区间为[π\6+kπ，2π\3+kπ]
