已知6sin^2θ+sinθcosθ-2cos^2θ=0,θ∈[π/2,π]求cos2θ的值
问题描述：
已知6sin^2θ+sinθcosθ-2cos^2θ=0,θ∈[π/2,π]求cos2θ的值
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)...