Question

已知函数f（x）=sin²x+根号3sinxcosx+2cos²x x∈R 求函数最值，周期，单调区间

Answer 1

f（x）=sin²x+根号3sinxcosx+2cos²x
=sin²x+cos²x+√3sinxcosx+cos²x
=1+(√3/2)sin2x+(1/2)cos2x+1/2
=3/2+sin2xcosπ/6+cos2xsinπ/6
=3/2+sin(2x+π/6)
所以：
函数的最大值=3/2+1=5/2
函数的最小值=3/2-1=1/2
最小正周期=2π/2=π
周期=kπ；k∈Z
单调增区间：
2kπ-π/2<2x+π/6<2kπ+π/2
2kπ-2π/3<2x<2kπ+π/3
kπ-π/3<x<kπ+π/6；k∈Z
单调减区间：
2kπ+π/2<2x+π/6<2kπ+3π/2
2kπ+π/3<2x<2kπ+4π/3
kπ+π/6<x<kπ+2π/3；k∈Z

Answer 2

f（x）=sin²x+根号3sinxcosx+2cos²x
=(1-coss2x)/2+√3/2sin2x+(1+cos2x)
=√3/2sin2x+1/2cos2x+3/2
=sin(2x+π/6)+3/2
函数最大值=1+3/2=5/2
函数最小值=-1+3/2=1/2
周期T=2π/2=π

2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
增区间为 【kπ-π/3，kπ+π/6】 （k∈Z）
2kπ+π/2<=2x+π/6<=2kπ+3π/2
kπ+π/6<=x<=kπ+2π/3
增区间为 【kπ+π/6，kπ+2π/3】 （k∈Z）