设α是第三象限角,且sinα^4+cosα^4=5/9,求tanα

α是第三象限角,tanα＞0
sinα^4+cosα^4=5/9
(sinα^2+cosα^2)^2=1
sinα^4+cosα^4+2sinα^2cosα^2=1
2sinα^2cosα^2=1-5/9=4/9
sinα^2cosα^2=2/9
sinα^2=tan^2α/(1+tan^2α)
cosα^2=1/(1+tan^2α)
sinα^2cosα^2=2/9
tan^2α/(1+tan^2α) * 1/(1+tan^2α) = 2/9
9tan^2α = 2tan^4α+4tan^2α+2
(2tan^4α-5tan^2α+2)=0
(2tan^2α-1)(tan^2α-2)=0
2tan^2α=1,或tan^2α=2
tanα＞0
tanα=根号2/2,或tanα=根号2

sinα^4+cosα^4=5/9
sinα^4+2sinα^2cos^2α+cosα^4-2sinα^2cos^2α=5/9
(sinα^2+cos^2α)^2-2sinα^2cos^2α=5/9
1-2(1-sinα^2)sin^2α=5/9
1-2sinα^2+2sin^4α=5/9
sinα^4-sinα^2+2/9=0
(sinα^2-1/...