已知函数f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx,求函数f（x）的最大值、最

f(x)=2cosx*sin(x+π/3)-√3sin²x+sinxcosx
=[sin(2x+π/3)+sin(π/3)]-[(√3/2)*(2sin²x-1)+√3/2]+(1/2)*sin2x
=sin(2x+π/3)+[sin(π/3)-√3/2]-(√3/2)*(-cos2x)+(1/2)*sin2x
=sin(2x+π/3)+(√3/2)*cos2x+(1/2)*sin2x
=2sin(2x+π/3)
当2x+π/3=kπ+π/2(k∈z),即x=kπ/2+π/12时(k∈z),f(x)有最大值,最大值为2
当2x+π/3=kπ-π/2(k∈z),即x=kπ/2-5π/12时(k∈z),f(x)有最小值,最小值为-2

f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx
=2cosx[1/2sinx+√3/2cosx]-√3sin²x+sinxcosx
=√3cos²x-√3sin²x+2sinxcosx
=√3cos2x+sin2x
=2sin(2x+派/3）
剩下的自己做就行了

f(x)=2cosx[(1/2)sinx+(√3/2)cosx]-√3sin²x+sinxcosx+1
=cosxsinx+√3cos²x-√3sin²x+sinxcosx+1
=2sinxcosx+√3(cos²x-sin²x)+1
=sin(2x)+√3cos(2x)+1
=2[(1/2)sin(2x)+(√3/2)cos(2x)]+1
=2sin(2x+π/3)+1
所以，函数f(x)的最大值为3，最小正周期为π。