解
y=sin²x+2√3sinxcosx-cos²x
=√3sin2x-(cos²x-sin²x)
=√3sin2x-cos2x
=√(√3)²+1sin(2x-π/6)——公式asinx+bcosx=√(a²+b²)sin(x+t)(tant=b/a
=2sin(2x-π/6)
当2x-π/6=π/2+2kπ时,
即x=π/3+kπ时,y取得最大值
∴ymax=2
∴x的集合{x/x=π/3+kπ,k∈z}
当-π/2+2kπ≤2x-π/6≤π/2+2kπ
即-π/6+kπ≤x≤π/3+kπ时,
y是增函数
∴增区间[kπ-π/6,kπ+π/3]