求函數y=(sinx-2)/(sinx-1)的值域
ysinx-y=sinx-2,故得sinx=(y-2)/(y-1)；
(y-2)/(y-1)≦1.(1)
(y-2)/(y-1)≧-1.(2)
由(1)得(y-2)/(y-1)-1=(-1)/(y-1)≦0,即有1/(y-1)≧0,故得y>1.①
由(2)得(y-2)/(y-1)+1=(2y-3)/(y-1)≧0,故得y