当-π/2≤x≤π/2时,函数f(x)=sin(x-2π)+根号3cos(2π-x)的最大值和最小值分别是.
问题描述：
当-π/2≤x≤π/2时,函数f(x)=sin(x-2π)+根号3cos(2π-x)的最大值和最小值分别是.
答
f(x)=sin(x-2π)+√3cos(2π-x)
=2[(1/2)sinx+(√3/2)cosx]
=2sin(x+π/3)
∵-π/2≤x≤π/2
∴f(x)min=f(-π/2)=-1
f(x)max=f(π/6)=2