Question

已知函数f（x）=2sin²x+sin2x

求函数f（x）的最小正周期和最大值

Answer 1

f（x）=2sin²x+sin2x
=-(1-2sin²x)+sin2x+1
=sin2x-cos2x+1
==√2*(√2/2*sin2x-√2/2*cos2x)+1
=√2sin(2x-π/4)+1
最小正周期为：T=2π/2=π
∵-1≤sin(2x-π/4)≤1
∴1-√2≤√2sin(2x-π/4)+1≤1+√2
即f(x)的值域为：[1-√2，1+√2]
所以最大值为：1+√2。

Answer 2

答：
f(x)=2sin²x+sin2x
=1-cos2x+sin2x
=√2*[(√2/2)sin2x-(√2/2)cos2x]+1
=√2sin(2x-π/4)+1
最小正周期T=2π/2=π
最大值为√2+1