f（x）=sinxcosx+sin²x=(1/2)sin2x+(1/2)(1-cos2x)=(1/2)(sin2x-cos2x)+1/2=(√2/2)sin(2x-π/4)+1/2x∈[0 2]2x∈[0,4]所以,当2x=0；即x=0时；函数有最小值=0当2x=3π/4；即x=3π/8时；函数有最大值=(√2+1)/2...