f(x)=2cosx(sinx-cosx)+2=2cosx*sin x-2(cosx)^2+2=sin2x-[2(cosx)^2-1]+1=sin2x-cos2x+1=根号(2)*sin(2x-π/4)+11、T=2π/2=π2、x∈[π/8,3π/4],2x-π/4∈[0,5π/4],所以sin(2x-π/4)∈[-根号(2)/2,1]f(x))∈[0,1...