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已知6sin^2θ+sinθcosθ-2cos^2θ=0,θ∈[π/2,π]求cos2θ的值
人气:480 ℃ 时间:2020-08-13 06:24:19
解答
6sin^2θ+sinθcosθ-2cos^2θ=0两边同时除以cos^2θ6tan^2θ+tanθ-2=0(2tanθ-1)(3tanθ+2)=0θ∈[π/2,π]tanθ=-2/3cos2θ=(cos^2θ-sin^2θ)/(cos^2θ+sin^2θ)=(1-tan^2θ)/(1+tan^2θ)=(1-4/9)/(1+4/9)=(5/9)...
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