已知sinθ+cosθ=1/5,θ∈(o,π),求cos(θ-π/3)+cotθ

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sinθ=1/5-cosθ;
(sinθ)^2=1/25-(2/5)cosθ+(cosθ)^2;
1-(cosθ)^2=1/25-(2/5)cosθ+(cosθ)^2;
2(cosθ)^2-(2/5)cosθ-24/25=0;
25(cosθ)^2-5cosθ-12=0;
cosθ=4/5或-3/5;
cos(θ-π/3)+cotθ
=cosθcosπ/3+sinθsinπ/3+cosθ/sinθ
=cosθ/2+(√3/2)sinθ+cosθ/sinθ
=cosθ/2+(√3/2)(1/5-cosθ)+cosθ/(1/5-cosθ)
当cosθ=4/5时,cosθ/2+(√3/2)(1/5-cosθ)+cosθ/(1/5-cosθ)=2/5-(√3/2)(3/5)-4/3;
当cosθ=-3/5时,cosθ/2+(√3/2)(1/5-cosθ)+cosθ/(1/5-cosθ)=-3/10+2√3/5-3/4

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