2sinπ/6x=1-cosπ/3x
f（x）=2sinπ/6x+sin（π/3x+π/6）—1
=（1-cosπ/3x)+sinπ/3xcosπ/6+cosπ/3xsinπ/6-1
=√3/2sinπx/3-1/2cosπx/3
=sin(πx/3-π/6)
∴f(x)max=1,f(x)min=-1
∴f(x)的值域为[-1,1]希望能帮到你，不懂可以问！