1.
f(x)=sin²x+2sinxcosx+3cos²x+m
=2sinxcosx+2cos²x+sin²x+cos²x+m
=sin(2x)+cos(2x)+1+1+m
=sin(2x)+cos(2x)+m+2
=√2sin(2x+π/4) +m+2
2kπ-π/2≤2x+π/4≤2kπ+π/2 (k∈Z)时,函数单调递增,此时
kπ-3π/8≤x≤kπ+π/8 (k∈Z)
函数的单调递增区间为[kπ-3π/8,kπ+π/8] (k∈Z)
2.
sin(2x+π/4)=1时,f(x)有最大值[f(x)]max=m+2+√2；
sin(2x+π/4)=-1时,f(x)有最小值[f(x)]min=m+2-√2.
函数值域为[m+2-√2,m+2+√2],又已知函数值域为[2-√2,2+√2],因此
m+2-√2=2-√2
m+2+√2=2+√2
解得m=0.