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Question Number 41902 by upadhyayrakhi20@gmail.com last updated on 15/Aug/18

tan3θ tan2θ =1  find the general^(solution......)

tan3θtan2θ=1findthegeneralsolution......

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

tan3θtan2θ1=02cos5θcos5θ+cosθ=0(sorrynotimefortypingatthemoment)cos5θ=0cos5θ+cosθ0θ=±π10+πzθ=±3π10+πz;zZ

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