证明:由已知得3sin2α=1-2sin2β=cos2β,
又sin2β=sin2α=3sinαcosα,
∴cos(α+2β)=cosαcos2β-sinαsin2β
=cosα3sin2α-sinα3sinαcosα=0.
又0<α<,0<β<,
∴0<α+2β<π.∴α+2β=.