Question

已知函数F(X)=sin²X+√3sinxcosx+2cos²X,X∈R 1 . 求函数的最小正周期 和 单调递增区间

Answer 1

F(X)=sin²X+√3sinxcosx+2cos²X
=1/2(1-cos2x)+√3/2*sin2x+1+cos2x
=√3/2sin2x+1/2cos2x+3/2
=sin(2x+π/6)+3/2
函数的最小正周期T=2π/2=π
由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z
得2kπ-2π/3≤2x≤2kπ+π/3,k∈Z
∴kπ-π/3≤x≤kπ+π/6,k∈Z
∴函数递增区间为
[kπ-π/3,kπ+π/6],k∈Z

Answer 2

F(X)=sin²X+√3sinxcosx+2cos²X
=（1-cos2X）/2 + √3/2 sin2X + 1+cos2X
=√3/2 sin2X + 1/2cos2X + 3/2
=sin(2X+π/6) +3/2

最小正周期T=2π/2=π

单调递增区间 (k∈Z)
2X+π/6∈[2kπ-π/2，2kπ+π/2]
2X∈[2kπ-π2/3，2kπ+π/3]
X∈[kπ-π/3，kπ+π/6]