奇函数f(x)的定义域为R所以f(0)=0f(cos2x-3)+f(4m-2mcosx)>0f(cos2x-3)>-f(4m-2mcosx)f(cos2x-3)>f(-4m+2mcosx)即cos2x-3>-4m+2mcosx2(cosx)^2-2mcosx+4m-4>00≤cosx=t≤12t^2-2mt+4m-4>0m>(t2-2)/(t-2)又(t2-2)/(t-2)=4-[(2-t)+2/(2-t)]≤4-2√2所以m>4-2√2