f(x)=3sin(x+π/3)+cos(x+π/6)
=3(sinx*cosπ/3+cosx*sinπ/3)+(cosx*cosπ/6-sinx*sinπ/6)
=(3/2)sinx+(3√3/2)cosx+(√3/2)cosx-(1/2)sinx
=sinx+2√3cosx
=√[1^2+(2√3)^2]*sin(x+y)
=√13sin(x+y),其中tany=2√3/1=2√3
所以T=2π/1=2π
sin最值±1
所以f(x)最大值=√13
最小值=-√13