Question

高中数学：已知函数f（x）=sin²x+2sinxcosx+3cos²x，x∈R

已知函数f（x）=sin²x+2sinxcosx+3cos²x，x∈RR，求：
（1）函数f（x）的最小正周期、最大值
（2）函数f（x）的单调增区间．

Answer 1

f(x)=1+2cos^2x+sin2x
=(cos2x+2)+sin2x
=(cos2x+sin2x）+2
=√2sin(2x+π/4）+2
(1)函数f（x）的最小正周期为π、最大值为√2+2；
(2)2kπ-π/2<=2x+π/4<2kπ+π/2时单调递增，即
kπ-3π/8<=x<kπ+π/8为单调增区间 或[kπ-3π/8，kπ+π/8）
2kπ+π/2<=2x+π/4<2kπ+3π/2时单调递减，即
kπ+π/8<=x<kπ+5π/8为单调减区间 或[kπ+π/8，kπ+5π/8）
k为整数。

Answer 2

f（x）=sin²x+2sinxcosx+3cos²x
=sin2x+2cos²x+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
(1) 最小正周期=2π/2=π
f(x)最大=2+√2
(2) 单增区间2x+π/4∈[2kπ-π/2, 2kπ+π/2]
x∈[kπ-3π/8, kπ+π/8] k∈Z
希望能帮到你O(∩_∩)O

Answer 3

fx=1-2sinx^2+sin2x=cos2x+sin2x=根号2sin(2x+45)+2
1.t=pai,最大值2+根号2
2.2x+45在(-π/2+2kπ，π/2+2kπ）解处即可