y=(sinx+cosx)²+2cos²x
=sin²x+cos²x+2sinxcosx+2cos²x
=1+sin2x+cos2x-1
=sin2x+cos2x
=√2(sin2x·√2/2+√2/2·cos2x)
=√2sin(2x+π/4)
x∈[0,π/2]
0≤2x≤π
π/4≤2x+π/4≤5π/4
所以
當(dāng)2x+π/4=π/2時(shí),sin(2x+π/4)=1,取得最大值√2
當(dāng)2x+π/4=5π/4時(shí),sin(2x+π/4)=-√2/2,取得最小值-√2/2×√2=-1
值域?yàn)閇-1,√2]不是=1+sin2x+cos2x+1嗎？cos2x=2cos²x-1

2cos²=cos2x+1
對(duì)，不好意思，我搞錯(cuò)1的符號(hào)了

y=(sinx+cosx)²+2cos²x
=sin²x+cos²x+2sinxcosx+2cos²x
=1+sin2x+cos2x+1
=sin2x+cos2x+2
=√2(sin2x·√2/2+√2/2·cos2x)+2
=√2sin(2x+π/4)+2
x∈[0,π/2]
0≤2x≤π
π/4≤2x+π/4≤5π/4
所以
當(dāng)2x+π/4=π/2時(shí)，sin(2x+π/4)=1，取得最大值2+√2
當(dāng)2x+π/4=5π/4時(shí)，sin(2x+π/4)=-√2/2，取得最小值2-√2/2×√2=2-1=1
值域?yàn)閇1，2+√2]