α,β∈(0 ,∏/2),且3(sinα)^2 +2(sinβ)^2=1 ,3sin2α-2sin2β=0.
问题描述：
α,β∈(0 ,∏/2),且3(sinα)^2 +2(sinβ)^2=1 ,3sin2α-2sin2β=0.
求α+2β的值.
答
解法一:
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=1
9(sinα)^2[(sinα)^2+(cosα)^2]=1
9(sinα)^2=1所以(sinα)^2=1/9
因为0