1、
f(x)=sin(2x-π/6)-2cos²x
=(√3/2)sin2x-(1/2)cos2x-(1+cos2x)
=(√3/2)sin2x-(3/2)cos2x-1
=√3sin(2x-π/3)-1
最小正周期T=2π/2=π
2、
x∈[-π/6,π/3]
则：2x-π/3∈[-2π/3,π/3]
则：sin(2x-π/3)∈[-1,√3/2]
所以,f(x)∈[-√3-1,1/2]
当2x-π/3=π/3,即：x=π/3时,f(x)有最大值1/2