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Thread: trigo and log equation

  1. #1
    Super Member dhiab's Avatar
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    trigo and log equation

    Solve in R :
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by dhiab View Post
    Solve in R :
    I will try

    let $\displaystyle t=\pi (ln(x^2))$

    [tex]sint + cost = 1 [tex]

    $\displaystyle 2sin(t)cos(t) + 1 - 2sin^2(t) = 1 $

    $\displaystyle 2sin(t)(cos(t) - sin(t)) = 0 $

    $\displaystyle sint=0........ln(x^2)=0.....x=\pm 1 $

    $\displaystyle sint=cost....t=\frac{\pi}{4} + 2\pi n .....$

    $\displaystyle \pi ln(x^2)=\pi ( \frac{1}{4} + 2n )$

    $\displaystyle ln(x^2)= \frac{1}{4} + 2n $

    $\displaystyle x^2 = e^{.75 + 2n } $

    $\displaystyle x= e^{.375 + n } $
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by Amer View Post
    I will try

    let $\displaystyle t=\pi (ln(x^2))$

    [tex]sint + cost = 1 [tex]

    $\displaystyle 2sin(t)cos(t) + 1 - 2sin^2(t) = 1 $

    $\displaystyle 2sin(t)(cos(t) - sin(t)) = 0 $

    $\displaystyle sint=0........ln(x^2)=0.....x=\pm 1 $

    $\displaystyle sint=cost....t=\frac{\pi}{4} + 2\pi n .....$

    $\displaystyle \pi ln(x^2)=\pi ( \frac{1}{4} + 2n )$

    $\displaystyle ln(x^2)= \frac{1}{4} + 2n $

    $\displaystyle x^2 = e^{.75 + 2n } $

    $\displaystyle x= e^{.375 + n } $
    Hello Amer
    here is expression just : sint = 2sin(t/2)cos(t/2) and :
    cost = 1-2sin2 (t/2)
    TANK YOU
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  4. #4
    Senior Member pankaj's Avatar
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    Quote Originally Posted by dhiab View Post
    Solve in R :
    Divide by $\displaystyle \sqrt{2}$ on both sides

    $\displaystyle \sin \frac{\pi}{4} \sin \pi ln(x^2)+\cos \frac{\pi}{4} \cos \pi ln(x^2)=\frac{1}{\sqrt{2}}$

    $\displaystyle
    \cos(\frac{\pi}{4}+\pi ln(x^2))=\cos\frac{\pi}{4}
    $

    $\displaystyle
    \frac{\pi}{4}+\pi ln(x^2)=2n\pi \pm \frac{\pi}{4}
    $

    $\displaystyle ln(x^2)=2n$;$\displaystyle x^2=e^{2n}$;$\displaystyle x=e^{n}$

    $\displaystyle ln(x^2)=\frac{1}{2}-2n$;$\displaystyle x^2=e^{\frac{1}{2}-2n}$;$\displaystyle x=e^{\frac{1}{4}-n}$

    'n' being any integer
    Last edited by pankaj; Jun 9th 2009 at 08:29 AM.
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by pankaj View Post
    Divide by $\displaystyle \sqrt{2}$ on both sides

    $\displaystyle \sin \frac{\pi}{4} \sin \pi ln(x^2)+\cos \frac{\pi}{4} \cos \pi ln(x^2)=\frac{1}{\sqrt{2}}$

    $\displaystyle
    \cos(\frac{\pi}{4}+\pi ln(x^2))=\cos\frac{\pi}{4}
    $

    $\displaystyle
    \frac{\pi}{4}+\pi ln(x^2)=2n\pi \pm \frac{\pi}{4}
    $

    $\displaystyle ln(x^2)=-2n$;$\displaystyle x^2=e^{-2n}$;$\displaystyle x=e^{-n}$

    $\displaystyle ln(x^2)=\frac{1}{2}-2n$;$\displaystyle x^2=e^{\frac{1}{2}-2n}$;$\displaystyle x=e^{\frac{1}{4}-n}$

    'n' being any integer
    Hello : i' cannot understing this resolution :
    Thank you
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  6. #6
    Senior Member pankaj's Avatar
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    I made a mistake.I have edited my post.
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