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Question-122505




Question Number 122505 by aupo14 last updated on 17/Nov/20
Answered by MJS_new last updated on 17/Nov/20
t=tan (θ/2)  5((2t)/(t^2 +1))+((t^2 −1)/(t^2 +1))=((t^2 +1)/(2t))  t^4 −t^3 −8t^2 +t+1=0  (t^2 −3t−1)(t^2 +2t−1)=0  t=−1±(√2)∨t=((3±(√(13)))/2)  ⇒  θ=2nπ+(π/4)∨θ=2nπ−((3π)/4)∨θ=2nπ+2arctan ((3±(√(13)))/2)
$${t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{5}\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{t}} \\ $$$${t}^{\mathrm{4}} −{t}^{\mathrm{3}} −\mathrm{8}{t}^{\mathrm{2}} +{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=−\mathrm{1}\pm\sqrt{\mathrm{2}}\vee{t}=\frac{\mathrm{3}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\theta=\mathrm{2}{n}\pi+\frac{\pi}{\mathrm{4}}\vee\theta=\mathrm{2}{n}\pi−\frac{\mathrm{3}\pi}{\mathrm{4}}\vee\theta=\mathrm{2}{n}\pi+\mathrm{2arctan}\:\frac{\mathrm{3}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$

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