已知函数f（x）=4sin^2x+2sin2x-2求fx的最小正周期 f（x）的最大值及此时x的集合（2）证明函数f（x

(1)f(x)=4sin^2x+2sin2x-2
=2-2cos2x+2sin2x-2
=2sin2x-2cos2x
=2√2sin(2x-π/4)
T=2π/w=2π/2=π
2x-π/4=2kπ+π/2
x=kπ+3π/8,k属于z
f(x)max=2√2
(2)f(x)=2√2sin(2x-π/4)
f(-x-π/4)=2√2sin[2(-x-π/4)-π/4]
=2√2sin(-2x-3π/4)
=-2√2sin(2x+3π/4)
=2√2sin(2x-π/4)
即f(x)=f(-x-π/4),
即f(x)的图像关于直线x=-π/8对称

f(x)=2sin2x-2(1-2sin8x)
=2sin2x-2cos2x
=2√2sin(2x-π/4)
所以T=2π/2=π
2x-π/4=2kπ+π/2
{x|x=kπ+3π/8},最大是2√2
f(-π/8)=2√2sin(-π/2)=-2√2
取到最小值，所以x=-π/8是对称轴