已知函数f(x)=sin^2(π/4+x)+cos^x+1/2,x∈R

(1)f(x)=sin^2(π/4+x)+cos^x+1/2
=[1-cos（π/2+2x)]/2+(1+cos2x)/2+1/2
=3/2+(sin2x+cos2x)/2
=3/2+√2sin(2x+π/4)/2
f(x）的最大值为(3+√2)/2,最小值为(3-√2)/2
最小正周期T=π
(2)f（x）≥3/2,即
3/2+√2sin(2x+π/4)/2≥3/2,即
sin(2x+π/4)≥0
2kπ≤2x+π/4≤2kπ+π
kπ-π/8≤x≤kπ+5π/8,又x∈[0,π]
所以7π/8≤x≤π

太吝啬了 没分