已知3sin^2α+sin^2β=2sinα,求sin^2α+sin^2β的最大值

希望LZ有耐心看下去：
tan(α+β)=2tan(α-β)
tan(α+β)/tan(α-β)=2
[sin(α+β)/cos(α+β)]/[sin(α-β)/cos(α-β)]=2
[sin(α+β)/cos(α+β)]*[cos(α-β)/sin(α-β)]=2
[(sinαcosβ+cosαsinβ)*(cosαcosβ+sinαsinβ)]/
[(cosαcosβ-sinαsinβ)*(sinαcosβ-cosαsinβ)]=2
[sinαcosαcos^2(β)+sin^2(α)sinβcosβ+cos^2(β)sinβcosβ+
sinαcosαsin^2(β)]/[sinαcosαcos^2(β)-cos^2(α)sinβcosβ-
sin^2(α)sinβcosβ+sinαcosαsin^2(β)]=2
/=2
(sinαcosα+sinβcosβ)/(sinαcosα-sinβcosβ)=2
(sin2α/2+sin2β/2)/(sin2α/2-sin2β/2)=2
(sin2α+sin2β)/(sin2α-sin2β)=2
sin2α+sin2β=2sin2α-2sin2β
3sin2β=sin2α
sin2β/sin2α=1/3

3sin^2α+sin^2β=2sinα,sin^2β=2sinα-3sin^2α,所以sin^2α+sin^2β=sin^2α+2sinα-3sin^2α=-2sin^2α+2sinα,令sinα=t,t∈[-1,1]-2t^2+2t=-2(t-1/2)^2+1/2对称轴t=1/2在区间内,所以t=1/2时取最大值1/2,所以s...