f(x)=3sin(x+π/3)+√3cos[π/2-(π/6-x)]
=3sin(x+π/3)+√3cos(x+π/3)
=2√3[√3/2*sin(x+π/3)+1/2*cos(x+π/3)]
=2√3[sin(x+π/3)cosπ/6+cos(x+π/3)sinπ/6]
=2√3sin(x+π/3+π/6)
sin(x+π/3+π/6)最大=1
所以f(x)最大=2√3