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Thread: Solving trig equations with involving aperiodic functions

  1. #1
    Super Member maxpancho's Avatar
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    Question Solving trig equations with involving aperiodic functions

    Solving trig equations with involving aperiodic functions-problem201.png

    C)

    Principal domain:

    $-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}$

    $-1.41099 \leq x \leq 0.91099$

    Next:

    $\sin(2x^2+x-1)=\dfrac{2}{5}$

    $2x^2+x-1=\arcsin(\dfrac{2}{5})$

    $x_1=-1.1265; x_2=0.62650$

    Since $\sin(\theta)=-\sin(\theta)+\pi$ we get:

    $\sin(-(2x^2+x-1)+\pi)=\dfrac{2}{5}$

    $-2x^2-x+1+\pi=\arcsin(\dfrac{2}{5})$

    Hence:

    $x_3=1.1384; x_4=-1.6384$

    Although, I wouldn't be able to find any more values of x by the methods I have been introduced to, since the functions is aperiodic.


    By the same method I can find two values of x in D), but that's it. Although, I think they want me to do something else, rather than values for x.



    P.S. Also, I haven't been introduced to solving trig equations with aperiodic functions, so it must be that they want to stretch me a little. So the tasks are probably rigged somehow, like in the case of C) I can find exactly four values with the methods I've been shown (well, as I understand), and that's what they want from me in the problem.

    So, I guess it must be similar with the D) task, but I can't figure it out.
    Last edited by maxpancho; Jul 13th 2014 at 09:44 AM.
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  2. #2
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    Re: Solving trig equations with involving aperiodic functions

    Where did your "principal domain" come from? I think it would more likely be -\pi\leq 2x^2+x-1\leq \pi.

    Then knowing that tan (x) = tan (x + \pi) you should be able to find four solutions.
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