因为,对于sinx,其周期是2π,下面仅在x∈[0,2π]内考虑问题.
y=(1-2sinx)/(1+3sinx)
显然：sinx≠-1/3,即：x≠arcsin(-1/3)
y'=[-2cosx(1+3sinx)-3cosx(1-2sinx)]/(1+3sinx)²
y'=(-2cosx-6sinxcosx-3cosx+6sinxcosx)/(1+3sinx)²
y'=-5cosx/(1+3sinx)²
1、令：y'＞0,即：-5cosx/(1+3sinx)²＞0
有：5cosx＜0
解得：cosx＜0
即：x∈(π/2,arcsin(-1/3))∪(arcsin(-1/3),3π/2)时,y为单调增函数.
2、令：y'＜0,即：-5cosx/(1+3sinx)²＜0
有：5cosx＞0
即：x∈[0,π/2)∪(3π/2,arcsin(-1/3))∪(arcsin(-1/3),2π]时,y为单调减函数.
综合以上,有：
①x=π/2时：y有极小值,y极小=[1-2sin(π/2)]/[1+3sin(π/2)]=-1/4
②x=3π/2时：y有极大值,y极大=[1-2sin(3π/2)]/[1+3sin(3π/2)]=-3/2
③x=arcsin(-1/3)时：y=[1-2sin(arcsin(-1/3)]/[1+3sin(arcsin(-1/3)]=∞
④x=0时：y=(1-2sin0)/(1+3sin0)=1
因此,所求值域为：y∈[-3/2,∞)