一道三角函数题求解!设函数f(x)=2sin^2(wx+派/4)+2cos^2wx(w>1)的图像上两个相邻的最低点之间

(1)f(x)=2sin²(wx+π/4)+2cos²wx (w>1)
=1-cos(2wx+π/2)+1+cos(2wx)
=2+sin(2wx)+ cos(2wx)
=2+√2sin(2wx+π/4)
T=2π/|2w|=2π/3
w=±3/2,w>1,w=3/2,
f(x)= 2+√2sin(3x+π/4)
当sin(3x+π/4)=1时,f(x)有最大值f（x）max=2+√2,
此时3x+π/4=2kπ+π/2,解得x=(2kπ+π/4)/3=2kπ/3+π/12, k∈N;
(2) y=f(x)的图像向右平移π/8个单位长度,
y=f(x-π/8)= 2+√2sin[3(x-π/8)+π/4]=2+√2sin(3x-π/8),
沿y轴对称后,g(x)=2+√2sin(-3x-π/8)=2-√2sin(3x+π/8),
设t= sin(3x+π/8), 2-√2t为减函数,复合函数的单调性可得,
2kπ-π/2≤3x+π/8≤2kπ+π/2时,t是关于x的增函数,g(x)
是关于x的减函数,
g(x)的单调减区间为[2kπ/3-5π/24, 2kπ/3+π/8], k∈N