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Thread: Trig equation help!!

  1. #1
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    Trig equation help!!

    Hi guys here is my question cos3x= sinx for the domain--- 0</ x </ Pi/2

    any help would be appreciated
    Last edited by Gurp925; Oct 25th 2013 at 11:47 PM.
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    Re: Trig equation help!!

    Transform the equation like this.... sin(90-3x)=sinx then the rest is easy.............
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    Re: Trig equation help!!

    How did you get sin(90-3X) ?
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    Re: Trig equation help!!

    Quote Originally Posted by MINOANMAN View Post
    Transform the equation like this.... sin(90-3x)=sinx then the rest is easy.............
    No it's not still easy, as you have to take into account the symmetries of the unit circle as well.

    You're better off using the identity \displaystyle \begin{align*} \cos{(3\theta)} = 4\cos^3{(\theta)} - 3\cos{(\theta)} \end{align*}, then you have

    \displaystyle \begin{align*} \cos{(3x)} &= \sin{(x)} \\ 4\cos^3{(x)} - 3\cos{(x)} &= \sin{(x)} \\ \left[ 4\cos^3{(x)} - 3\cos{(x)} \right] ^2 &= \sin^2{(x)} \\ 16\cos^6{(x)} - 24\cos^4{(x)} + 9\cos^2{(x)} &= 1 - \cos^2{(x)} \\ 16\cos^6{(x)} - 24\cos^4{(x)} + 10\cos^2{(x)} - 1 &= 0 \\ 16X^3 - 24X^2 + 10X - 1 &= 0 \textrm{ if we let } X = \cos^2{(x)} \end{align*}

    Solve this cubic, use that to solve for x.
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    Re: Trig equation help!!

    I don't understand why some people complicate things to such extend that their advises become useless for the students.

    as I said cos3x=sin(90-3x) this is a formula of basic trigonometry no doubt about that.
    then the equation is transformed to sin(90-3x) =sinx
    this yields two formulas of general solutions..

    1. 90-3x =(360)κ +x

    2. 90-3x = (360)k+(180-x)

    where k is an arbitrary integer...that is to be chosen to fit your given interval.....

    no need to expand and get a qubic equation...........................
    Last edited by MINOANMAN; Oct 26th 2013 at 07:11 AM.
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