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tan3-tan2-1-find-the-general-solution-




Question Number 41902 by upadhyayrakhi20@gmail.com last updated on 15/Aug/18
tan3θ tan2θ =1  find the general^(solution......)
$$\mathrm{tan3}\theta\:\mathrm{tan2}\theta\:=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:^{\mathrm{solution}……} \\ $$
Answered by MJS last updated on 15/Aug/18
tan 3θ tan 2θ −1=0  −2((cos 5θ)/(cos 5θ +cos θ))=0  (sorry no time for typing at the moment)  cos 5θ =0 ∧ cos 5θ +cos θ ≠0  θ=±(π/(10))+πz ∨ θ=±((3π)/(10))+πz; z∈Z
$$\mathrm{tan}\:\mathrm{3}\theta\:\mathrm{tan}\:\mathrm{2}\theta\:−\mathrm{1}=\mathrm{0} \\ $$$$−\mathrm{2}\frac{\mathrm{cos}\:\mathrm{5}\theta}{\mathrm{cos}\:\mathrm{5}\theta\:+\mathrm{cos}\:\theta}=\mathrm{0} \\ $$$$\left(\mathrm{sorry}\:\mathrm{no}\:\mathrm{time}\:\mathrm{for}\:\mathrm{typing}\:\mathrm{at}\:\mathrm{the}\:\mathrm{moment}\right) \\ $$$$\mathrm{cos}\:\mathrm{5}\theta\:=\mathrm{0}\:\wedge\:\mathrm{cos}\:\mathrm{5}\theta\:+\mathrm{cos}\:\theta\:\neq\mathrm{0} \\ $$$$\theta=\pm\frac{\pi}{\mathrm{10}}+\pi{z}\:\vee\:\theta=\pm\frac{\mathrm{3}\pi}{\mathrm{10}}+\pi{z};\:{z}\in\mathbb{Z} \\ $$

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