3sin^2(α）+2sin^2(β)=2sinα则cos^2(α)+cos^2(β)的取值范围

3sin^2(α)+2sin^2(β)=2sinα
sin^2(α)+(2-2cos^2(α))+(2-2cos^2(β))=2sinα
cos^2(α)+cos^2(β)=[sin^2(α)-2sinα+4]/2
设sinα=t
因为2sin^2(β)=2sinα-3sin^2(α)≥0
2t-3t^2≥0
2/3≥t≥0
cos^2(α)+cos^2(β)=[t^2-2t+4]/2
=[(t-1)^2+3]/2
对称轴为t=1
所以取最大值时t=0 cos^2(α)+cos^2(β)=2
取最小值时t=2/3 cos^2(α)+cos^2(β)=14/9
所以取值范围[14/9,2]

3sin^2(α）+2sin^2(β)=2sinα
sin^2(α）+2-2cos^2(α)+2-2cos^2(β)=2sinα
cos^2(α)+cos^2(β)=[sin^2(α）+4-2sinα]/2=[(sinα-1)^2+3]/2
所以取值范围[3/2,7/2]

令cos^2(α)+cos^2(β)=P
3sin^2(α）+2sin^2(β)=2sinα
sin^2(α）+4-2P=2sinα
2P=(sinα+1)^2+3 在3到7的范围,又P<=2
所以P在1.5到2