已知sin平方2α+sin2αcosα-cos2α=1 α∈（0,90°） 求sinα tanα

∵sin²(2α)+sin(2α)cosα-cos(2α)=1
∴sin²(2α)+sin(2α)cosα-cos(2α)-1=0
==>sin²(2α)+sin(2α)cosα-2cos²α=0
==>[sin(2α)+2cosα][sin(2α)-cosα]=0
==>(2sinαcosα+2cosα)(2sinαcosα-cosα)=0
==>2cos²α(sinα+1)(2sinα-1)=0
∵α∈(0,90°)
∴cosα≠0,sinα+1≠0
∴2sinα-1=0 ==>sinα=1/2
==>α=30°
==>tanα=√3/3
故sinα=1/2,tanα=√3/3