若tan^2θ-tanθ-2=0,（θ∈（0,π））求sin^2θ+3sinθcosθ-2cos^2θ的值
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若tan^2θ-tanθ-2=0,（θ∈（0,π））求sin^2θ+3sinθcosθ-2cos^2θ的值
若tan^2θ-tanθ-2=0,（θ∈（0,π））求sin^2θ+3sinθcosθ-2cos^2θ的值
若tan^2θ-tanθ-2=0,（θ∈（0,π））求sin^2θ+3sinθcosθ-2cos^2θ的值
用A代替那个角
解之得：tanA=2或tanA=-1
sin^2A+3sinAcosA-2cos^2A=cos^2A(tan^2A-2+3tanA)=4tanAcos^2A
当tanA=2时,cos^2A=1/5
此时原式=8/5
当tanA=-1时,cos^2A=1/2
此时原式=-2
θ∈(0，π)
tan²θ - tanθ - 2 = 0
tanθ = -1 或 tanθ = 2
sinθ = √2/2 , cosθ = -√2/2 或 sinθ = 2√5/5 , cosθ = √5/5
(1) sinθ = √2/2 , cosθ = -√2/2
sin²θ + 3sinθcosθ - 2cos²θ =...
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θ∈(0，π)
tan²θ - tanθ - 2 = 0
tanθ = -1 或 tanθ = 2
sinθ = √2/2 , cosθ = -√2/2 或 sinθ = 2√5/5 , cosθ = √5/5
(1) sinθ = √2/2 , cosθ = -√2/2
sin²θ + 3sinθcosθ - 2cos²θ = -2
(2) sinθ = 2√5/5 , cosθ = √5/5
sin²θ + 3sinθcosθ - 2cos²θ = 8/5
收起
将sin^2x+cos^x=1代入，则原式变化为3(sin^x+sinxcosx)-2，然后变成3（sin^2x+sinxcosx)/(sin^2x+cos^2x)-2,然后约去cos^2x,变成3(tan^2x+tanx)/(1+tan^2x)-2,变成只剩tanx的式子就好办了，（因为θ不好打，我改成x了）