求函数y=(2sinx-1)/(3sinx+2)在[π/2,3π/2)的值域

y=(2sinx+4/3-7/3)/(3sinx+2)
=(2sinx+4/3)/(3sinx+2)-(7/3)/(3sinx+2)
=2/3-7/(9sinx+6)
x范围则-1<=sinx<=1
-3<=9sinx+6<=15
1/(9sinx+6)<=-1/3,1/(9sinx+6)>=1/15
-7/(9sinx+6)>=7/3,-7/(9sinx+6)<=-7/15
值域是y<=1/5,y>=33sinx+2≠0→sinx≠-2/3没有考虑？x∈[π/2,3π/2)是否也没考虑？没错，采纳吧标准答案是y∈[-1/2,2/3),我也是百思不得其解，故请高人指点迷津。设y=1/3则sinx=5/3>1别问了，采纳吧