Question

函数y=sin^2x+2sinxcosx+3cos^2x 1求函数最小正周期 2求函数的单调递增区间 3当x取什么值 函数取最大值

Answer 1

解：y=sin²x+2sinxcosx+3cos²x
=sin²x+3cos²x+2sinxcosx
=2cos²x+1+sin2x
= =cos2x+sin2x
=√2sin(2x+π/4)
(1) 函数最小正周期为：2π/2=π
(2) 单调增区间为：[kπ-3π/8,kπ+π/8]
单调减区间为：[kπ+π/8,kπ+5π/8]
(3) 当x=kπ+π/8 时，函数取最大值√2。

Answer 2

y=sin^2x+2sinxcosx+3cos^2x
=sin^2x+3cos^2x+2sinxcosx
=2cos²+1+sin2x
=cos2x+2+sin2x
=√2sin(2x+π/4)+2
最小正周期为 2π/2=π
单调增区间
2x+π/4∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-3π/8,kπ+π/8]

单调减区间 [kπ+π/8,kπ+5π/8]

当取到最大值时
2x+π/4=2kπ+π/2
x=kπ+π/8