Question

已知函数fx=sin²x+√3sinxcosx+2cos²x,x∈R,求函数fx的最小周期和单调周

已知函数fx=sin²x+√3sinxcosx+2cos²x,x∈R,求函数fx的最小周期和单调周期

Answer 1

f(x)=sin²x+√3sinxcosx+2cos²x,
=√3sinxcosx+cos²x+1
=√3/2sin2x+1/2(1+cos2x)+1
=√3/2sin2x+1/2cos2x+3/2
=sin(2x+π/6)+3/2
∴f(x)最小正周期T=2π/2=π
由2kπ-π/2≤2x+π/6≤2kπ+π/2，k∈Z
得kπ-π/3≤x≤kπ+π/6，k∈Z
f(x)单调递增区间为[kπ-π/3,kπ+π/6]k∈Z

由2kπ+π/2≤2x+π/6≤2kπ+3π/2，k∈Z
得kπ+π/6≤x≤kπ+2π/3，k∈Z
f(x)单调递减区间为[kπ+π/6,kπ+2π/3]k∈Z

追问

谢谢

追答

OK