有哪位数学高手能帮我解一下这道题？请写出过程．谢谢！

Question

已知sinx+siny=2/3,求2/3+siny-cos^2x的取值范围.
问题补充：答案为[1/12,7/9]

Answer 1

2/3+siny-cos^2x=sin^2x+siny-1/3
=sin^2x-sinx+2/3-1/3
=sin^2x-sinx+1/3
=(sinx-1/2)^2+1/12,
因为-1<=siny<=1,则-1/3<=sinx<=1,
设sinx=z,-1/3<z<=1
f(z)=(z-1/2)^2+1/12>=1/12
当,-1/3<=z<1/2,f(z)为减函数f(z)<=f(-1/3)=7/9
当1/2<=z<=1,f(z)为增函数,f(z)<=f(1)=1/3
所以2/3+siny-cos^2x的取值范围为[1/12,7/9]

Answer 2

sinx+siny=2/3
sinx=2/3-siny
sin^2x=4/9+sin^2y-4siny/3
1-cos^2x=4/9+sin^2y-4siny/3
-cos^2x=sin^2y-4siny/3-5/9

siny-cos^2x=m
siny+sin^2y-4siny/3-5/9=m
sin^2y-siny/3-5/9-m=0
上方程未知数为siny的判别式
△=(-1/3)^2-4*1*(-5/9-m)≥0
m≥-7/12
siny-cos^2x≥-7/12
siny=2/3-sinx
2/3-sinx-cos^2x≥-7/12
2/3-sinx+1-cos^2x≥5/12
2/3-sinx+sin^2x≥5/12
sin^2x-sinx+1/4≥0
(sinx-1/2)^2≥0
sinx+siny=2/3
2/3≥six≥-1/3
8/9≥co^2x≥5/9
sinx=2/3-siny
2/3≥2/3-siny≥-1/3
1≥siny≥0
siny=1,sinx=-1/3,co^2x=8/9,siny-cos^2x=1-8/9=1/9
siny=0,sinx=2/3,co^2x=5/9.siny-cos^2x=0-5/9=-5/9
siny-cos^2x的最大值=1/9
siny-cos^2x≤1/9
故1/9≥siny-cos^2x≥-7/12
1/9+2/3≥2/3+siny-cos^2x≥2/3-7/12
7/9≥2/3+siny-cos^2x≥1/12