题目

3(sinα)^2+2(sinβ)^2=2sinα则(sinα)^2+(sinβ)^2取值范围^

提问时间：2021-03-14

答案

因为,3(sinα)^2+2(sinβ)^2=2sinα
所以,3(sinα)^2<=2sinα
所以,sinα(3sinα-2)<=0
则,0<=sinα<=2/3
(sinα)^2+(sinβ)^2=1/2[2(sinα)^2+2(sinβ)^2]
=(1/2)[3(sinα)^2+2(sinβ)^2-(sinα)^2]
=(1/2)[2sinα-(sinα)^2]
=(1/2)[1-(1-sinα)^2]
0<=sinα<=2/3
-2/3<=-sinα<=0
1/3<=1-sinα<=1
1/9<=(1-sinα)^2<=1
-1<=-(1-sinα)^2<=-1/9
0<=1-(1-sinα)^2<=8/9
0<=(1/2)[1-(1-sinα)^2]0<=4/9
则(sinα)^2+(sinβ)^2取值范围[0,4/9]