Question

已知y=sin²x+2根号3sinxcosx+3cos²x 求函数最小正周期和单调递增区间 过程谢谢

Answer 1

y=1-(cosx)^2+3(cosx)^2+√3sin2x
=1+2(cosx)^2+√3sin2x
=cos2x+√3sin2x+2
=2(sinπ/6cos2x+cosπ/6sin2x)+2
=2sin(2x+π/6）+2
最小正周期T=2π/2=π,
2kπ-π/2≤2x+π/6≤2kπ+π/2,为单调增函数，
kπ-π/3≤x≤kπ+π/6，k∈Z，
2kπ+π/2≤2x+π/6≤2kπ+3π/2,为单调减函数，
kπ+π/6≤x≤kπ+2π/3，k∈Z。