你好,你要的答案是:


f(x)=(1-cos2x)+sin(2x)*cos(π/6)+cos(2x)*sin(π/6)
=1+(√3/2)*sin2x-cos2x*(1/2)
=1+sin2x*cos(π/6)-cos2x*sin(π/6)
=1+sin(2x-π/6)
∴T=2π/2=π
∵X在区间[π/3, 5π/6],即(2x-π/6)在[π/2,3π/2]
∴ -1≤sin(2x-π/6)≤1
∴最大值=1+1=2
最小值=1-1=0