f(x)=sin²x+√3sinxcosx+2cos²x
=√3sinxcosx+cos²x+sin²x+cos²x
=√3/2sin2x+(1+cos2x)/2+1
=sin(2x+π/6)+3/2.
(1)当y取得最大值时,自变量x满足
2x+π/6=2kπ+π/2,x=kπ+π/6.
(2)T=2π/2=π;单调递增即
2kπ-π/2≤2x+π/6≤2kπ+π/2.
解得,x∈[kπ-π/3,kπ+π/6].