已知2sin2θ+1=cos2θ,求tanθ的值.

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已知2sin2θ+1=cos2θ,求tanθ的值.
已知2sin2θ+1=cos2θ,求tanθ的值.

已知2sin2θ+1=cos2θ,求tanθ的值.
2sin2θ+1=cos2θ
2sinθcosθ+1-cos2θ=0
2sinθcosθ+2sinθsinθ=0
2sinθ(cosθ+sinθ)=0
(1) sinθ=0
则tanθ=0
(2) cosθ+sinθ=0
tanθ=-1

4sinxcosx+(cosx)^2+(sinx)^2=(cosx)^2-(sinx)^2
令y=tanx
4y+1+y^2=1-y^2
y=0或-2

2sin2θ+1=cos2θ
2tan2θ+ sec2θ = 1
(2tan2θ-1)^2 = (sec2θ)^2
3(tan2θ)^2+ 4tan2θ = 0
tan2θ(3tan2θ+4 )=0
tan2θ = 0 or -3/4
case 1
tan2θ = 0
tanθ =0
case 2
tan2θ = ...

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2sin2θ+1=cos2θ
2tan2θ+ sec2θ = 1
(2tan2θ-1)^2 = (sec2θ)^2
3(tan2θ)^2+ 4tan2θ = 0
tan2θ(3tan2θ+4 )=0
tan2θ = 0 or -3/4
case 1
tan2θ = 0
tanθ =0
case 2
tan2θ = -3/4
2tanθ/(1-(tanθ)^2) = -3/4
8tanθ = -3+3(tanθ)^2
3(tanθ)^2 -8tanθ -3 =0
(3tanθ+1)(tanθ-3) =0
tanθ = -1/3 or 3

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2sin2θ+1=cos2θ
2sinθ*cosθ+1=1-2sin^2θ
2sinθ*cosθ=-2sin^2θ
cosθ=-sinθ
tanθ=-1

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