y=(sinx+1)/(cosx+2)
sinx+1=ycosx+2y
sinx-ycosx=2y-1
√(1+y^2)sin(x-φ)=(2y-1) (其中cosφ=1/√(1+y^2),sinφ=y/√(1+y^2) )
sin(x-φ)=(2y-1)/√(1+y^2)
|sin(x-φ)|=|2y-1|/√(1+y^2)≤1
|2y-1|≤√(1+y^2)
兩邊平方得：
4y^2-4y+1≤y^2+1
3y^2-4y≤0
y(3y-4)≤0
0≤y≤4/3
原函數(shù)的值域為：[0,4/3]