f(x)=3sin^2x+cos^2x-2sinx*cosx
=2sin^2x+sin^2x+cos^2x-2sinx*cosx
=1+2sin^2x-2sinxcox
=-cos2x-sin2x+2
=-√2sin(2x+π/4)+2
①求最小正周期 T=2π/2=π
②求单调减区间 2kπ-π/2≤2x+π/4≤2kπ+π/2
解得：kπ-3π/8≤x≤kπ+π/8
③当x∈[0,π/2]求函数的最大值
在此上为增函数所
当：：x=π/2时y有最大值为：y=3