y=sin2x+2√2cos(π/4+x）+3
=cos(2x-π/2)+2√2cos(π/4+x）+3
=1-2sin²(x-π/4)-2√2sin(x-π/4）+3
=4-2[sin²(x-π/4)+√2sin(x-π/4)]
令sin(x-π/4)=t,则-1≤t≤1于是
y=4-2(t²+√2t)
y=4-2(t²+√2t+1/2)+2*1/2
=4-2(t-√2/2)²+1
=5-2(t-√2/2)² （-1≤t≤1）
当t=-1时,函数取得最小值
4-2[(-1)²+√2*(-1)]
=4-2(1-√2)
=2-2√2
所以最小值为2-2√2