已知sin²2α+sin2αcos2α-cos2α=1,α∈(0,π)求sinα,tanα

问题描述:

已知sin²2α+sin2αcos2α-cos2α=1,α∈(0,π)求sinα,tanα

sin²2α+sin2αcos2α-cos2α=1,
(sin²2α-1)+cos2α(sin2α-1)=0,
(sin2α-1)(sin2α+1+cos2α)=0,
(1-sin2α)(2sinαcosα+2cos²α)=0,
2cosα(sinα-cosα)²(sinα+cosα)=0,
∴cosα=0,或sinα=cosα,或sinα= -cosα,
∵α∈(0,π),
∴α=π/2,或α=π/4,或α=3π/4,
当α=π/2时,sinα=1,tanα不存在;
当α=π/4时,sinα=√2/2,tanα=1;
当α=3π/4时,sinα= -√2/2,tanα= -1.