解由f（x)=sin(2x+π/3）-√3cos（2x+π/3）
=2(1/2sin(2x+π/3）-√3/2cos（2x+π/3）)
=2sin(2x+π/3-π/3)
=2sin2x
知當(dāng)2kπ-π/2≤2x≤2kπ+π/2,k屬于Z,y是增函數(shù)
即當(dāng)kπ-π/4≤x≤kπ+π/4,k屬于Z,y是增函數(shù)
故函數(shù)的增區(qū)間為[kπ-π/4,kπ+π/4],k屬于Z.
知當(dāng)2kπ+π/2≤2x≤2kπ+3π/2,k屬于Z,y是減函數(shù)
即當(dāng)kπ+π/4≤x≤kπ+3π/8,k屬于Z,y是減函數(shù)
故函數(shù)的減區(qū)間為[kπ+π/4,kπ+3π/8],k屬于Z.
由x屬于（-π/6,π/3）
知2x屬于（-π/3,2π/3）
即sin2x屬于(-√3/2,1]
即2sin2x屬于(-√3,2]
故函數(shù)
f（x）在（-π/6,π/3）上的值域(-√3,2].