求解一道数学高考题 谢谢

y=sinx/（2-cosx）
sinx=y(2-cosx)
1=sin^2x+cos^2x
=y^2(2-cosx)^2+cos^2x
=(y^2+1)cos^2x-4y^2cosx+4y^2

(y^2+1)cos^2x-4y^2cosx+4y^2-1=0
△=16y^4-4(y^2+1)(4y^2-1)=-12y^2+4≥0
y^2≤1/3
-√3/3≤y≤√3/3

两根和= 4y^2/(y^2+1)∈[-2,2]
2y^2≤2, y^2≤1,且
6y^2≥-2
满足

两根积=(4y^2-1)/(y^2+1)∈[-1,1]
(4y^2-1)/(y^2+1)≤1,3y^2≤2,y^2≤2/3
(4y^2-1)/(y^2+1)≥-1,5y^2≥0
满足

所以，函数y=sinx/（2-cosx）的值域是：[-√3/3,√3/3]

2y-ycosx=sinx
sinx+ycosx=2y
(1+y^2)^1/2[sinx*1/(1+y^2)^1/2+y/(1+y^2)^1/2 *cosx]=2y
(1+y^2)^1/2sin(x+a)=2y
-1<=sin(x+y)=2y/(1+y^2)^1/2<=1
-1<=2y/(1+y^2)^1/2<=1
解这个不等式组，得到的交集就是y的值域

点（2，，0）到点（COSX,-SINX)的斜率为Y，根据几何关系可求出范围（后者在单位圆上，做切线即可)