已知函数f(x)=2sinxcosx-2sinx²+1(x∈R)若在△ABC中,角ABC的对边分别为abc

f(x)=sin2x+cos2x,故：f(A+π/8)=sin(2A+π/4)+cos(2A+π/4)=sqrt(2)sin(2A+π/2)
=sqrt(2)cos(2A)=sqrt(2)/3,即：cos(2A)=1/3,故：1-2sinA^2=1/3,A是锐角
故：sinA=1/sqrt(3),而：2cosA^2-1=1/3,即：cosA=sqrt(2)/sqrt(3)
a^2=3=b^2+c^2-2bccosA≥2bc-(2sqrt(2)/sqrt(3))bc,故：bc≤3sqrt(3)/(2(sqrt(3)-sqrt(2)))
△ABC的面积S=(1/2)bcsinA≤(1/2)*(1/sqrt(3)*3sqrt(3)/(2(sqrt(3)-sqrt(2)))
=3(sqrt(3)+sqrt(2))/4,等号成立的条件：b=c=3(3+sqrt(6))/2