f(x)=cosx^4-2sinxcosx-sinx^4=cosx^4-sinx^2-2sinxcosx=(cosx^2+cosx^2)(cosx^2-sinx^2)-2sinxcosx=cos2x-sin2x=-根号2sin(2x-π/4)T=πx∈[0,π/2]所以2x-π/4∈[0,3π/4]所以当2x-π/4=0时最小,即x=π/8即f(x)min...