作业帮已知3sin^2α—2sinα+2sin^2β=0 求sin^2α+sin^2β的取值范围
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已知3sin^2α—2sinα+2sin^2β=0 求sin^2α+sin^2β的取值范围
已知3sin^2α—2sinα+2sin^2β=0 求sin^2α+sin^2β的取值范围
已知3sin^2α—2sinα+2sin^2β=0 求sin^2α+sin^2β的取值范围
可设a=sinα,b=sinβ,-1
因为sin^2b>=0,sin^2a>=0;
所以：2sin^2b+3sin^2a>=0.
即：2sina>=0,得到：sina>=0.
∵2sin^2b+3sin^2a=2sina，
∴sin^2b=sina-(3/2)sin^2a>=0
进一步：
Sina(1-3/2sina)>=0
1-3/2sina>=0,得到：...
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因为sin^2b>=0,sin^2a>=0;
所以：2sin^2b+3sin^2a>=0.
即：2sina>=0,得到：sina>=0.
∵2sin^2b+3sin^2a=2sina，
∴sin^2b=sina-(3/2)sin^2a>=0
进一步：
Sina(1-3/2sina)>=0
1-3/2sina>=0,得到：sina<=2/3.
即sina的取值范围为：[0,2/3].
则：
m=sin^2a+sin^2b
=sin^2a+sina-(3/2)sin^2a
=-(1/2)sin^2a+sina
=-(1/2)(sin^2a-2sina+1)+1/2
=-(1/2)(sina-1)^2+1/2.
因为0<=sina<=2/3.所以：
当sina=2/3，m有最大值m=4/9
当sina=0，m有最小值m=0。
所以m的取值范围为：[0,4/9].
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3sin^2α—2sinα+2sin^2β=0
→3sin^2a+2sin^b=2sina
→2sin^2a+2sin^b=2sina-sin^2a
→2*(sin^2a+sin^b)=-(sina-1)^2+1
∵-1<=sina<=1
∴-4<=-(sina-1)^2<=0
∴-3<=-(sina-1)^2+1<=1
∴-3<=2*(sin^2a+sin^b)<=1
即-3/2<=(sin^2a+sin^b)<=01/2