设0
由3(sinα)^2+2(sinβ)^2=1得3(sinα)^2+1-cos2β=1所以3(sinα)^2=cos2β(1)由3sin2α-2sin2β=0得3sinαcosα=sin2β(2)(1) 的平方+(2)的平方:9(sinα)^4+9(sinα)^2(cosα)^2=19(sinα)^2[(sinα)^2+(cosα)^2]=19(sinα)^2=1所以(sinα)^2=1/9因为0