Question

已知f(x)=sin^2x+2sinxcosx+3cos^2x,x∈(0,兀),求函数的最小正周期和值域;求函数的单调递增区间

Answer 1

2cos²x-1=cos2x
f(x)=sin²x+2sinxcosx+3cos²x
=(sin²x+cos²x)+(2cos²x-1)+1+2sinxcosx
=2+sin2x+cos2x
=2+(√2)sin[2x+(π/4)]
T=2π/做核ω=π
x∈(0,π)

[2x+(π/4)]∈(π/4,9π/4)
f(x)max=2+(√2)
f(x)min=2-(√2)
单调递并乱增区绝胡档间[2,kπ-π/2,2kπ+π/2]
2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-(3π/8)≤x≤kπ+(π/8)
根据题目条件
{x|0<x<π/8或5π/8<x<π}