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Math Help - Find the general solution for...

  1. #1
    Senior Member Educated's Avatar
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    Find the general solution for...

    I'm having a bit of trouble solving this trig problem:

    Find the General solution for... \sin(4x) - \cos(x) = 0
    At first when I looked at it it seemed easy but as I tried to solve it, it got really confusing and I got stuck on the \sin^3(x) part.

    Here's what I did:
    Double angle formula: \sin(2 \theta) = 2 \sin (\theta) \cos (\theta)
    Double angle formula: \cos(2 \theta) = 1-2\sin^2(\theta)

    0=\sin(4x) - \cos(x)
    0=2\cdot\sin(2x)\cdot\cos(2x) - \cos(x)
    0=4[sin(x)\cos(x)]\cdot[1-\sin^2(x)] - \cos(x)
    0=4\sin(x)\cos(x) - 8\sin^3(x)\cos(x) - \cos(x)

    0=\cos(x)[4\sin(x) - 8\sin^3(x) - 1]

    Now I can get the general solution for the \cos(x) part but what about the 4\sin(x) - 8\sin^3(x) - 1 part? How would I solve that?


    EDIT: Sorry I just realised I placed this in the wrong section :/
    Can a mod please move it to the trig section?
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  2. #2
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    Re: Find the general solution for...

    Quote Originally Posted by Educated View Post
    I'm having a bit of trouble solving this trig problem:



    At first when I looked at it it seemed easy but as I tried to solve it, it got really confusing and I got stuck on the \sin^3(x) part.

    Here's what I did:
    Double angle formula: \sin(2 \theta) = 2 \sin (\theta) \cos (\theta)
    Double angle formula: \cos(2 \theta) = 1-2\sin^2(\theta)

    0=\sin(4x) - \cos(x)
    0=2\cdot\sin(2x)\cdot\cos(2x) - \cos(x)
    0=4[sin(x)\cos(x)]\cdot[1-\sin^2(x)] - \cos(x)
    0=4\sin(x)\cos(x) - 8\sin^3(x)\cos(x) - \cos(x)

    0=\cos(x)[4\sin(x) - 8\sin^3(x) - 1]

    Now I can get the general solution for the \cos(x) part but what about the 4\sin(x) - 8\sin^3(x) - 1 part? How would I solve that?


    EDIT: Sorry I just realised I placed this in the wrong section :/
    Can a mod please move it to the trig section?
    You won't be able to solve \displaystyle 8\sin^3{x} - 4\sin{x} + 1 = 0 using analytical methods. You will need to use technology.
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  3. #3
    Senior Member BAdhi's Avatar
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    Re: Find the general solution for...

    Actually, \sin{x}=\frac{1}{2} seems to satisfy 8\sin^3{x}-4\sin{x}+1=0.
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    Re: Find the general solution for...

    Quote Originally Posted by BAdhi View Post
    Actually, \sin{x}=\frac{1}{2} seems to satisfy 8\sin^3{x}-4\sin{x}+1=0.
    Yes, that's true. However, if you are not a very good guesser, technology would be the way to go
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  5. #5
    Senior Member Educated's Avatar
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    Re: Find the general solution for...

    This was an old exam question so we're not allowed to use technology to solve it, just our brains.

    Although I guess you can guess one solution by trial and error for the cubic 8u^3 -4u +1 where u=\sin(x) and try to derive the other solutions from it by factoring.

    Thanks for the help!
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  6. #6
    Member sbhatnagar's Avatar
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    Re: Find the general solution for...

    \\8u^3-4u+1=0 \\ (2u-1)(4u^2+2u-1)=0\\u=\frac{1}{2},\hspace{10}\frac{1}{4}(\pm  \sqrt{5} -1)

    Therefore, \sin{x}= \frac{1}{2},\hspace{10}\frac{1}{4}(\sqrt{5}-1),\hspace{10}\frac{1}{4}(-\sqrt{5}-1).

    x=\arcsin{[\frac{1}{2}]}+2k\pi

    x=\arcsin{[\frac{1}{4}(\sqrt{5}-1)]}+2k\pi

    x=\arcsin{[\frac{1}{4}(-\sqrt{5}-1)]}+2k\pi

    where k is some integer.
    Last edited by sbhatnagar; November 5th 2011 at 03:39 AM.
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    Re: Find the general solution for...

    Hello, Educated!

    \text{Find the general solution for: }\:\sin(4x) - \cos(x) \:=\:0

    Here's a little help . . .

    Since \sin(4x) \,=\,\cos(x), then 4x and x are in the same right triangle.

    Code:
                  *
                 /|
                /x|
               /  |
              /   |
             /    |
            /     |
           /4x    |
          * - - - *
    Hence: . 4x + x \,=\,\tfrac{\pi}{2} \quad\Rightarrow\quad 5x \,=\,\tfrac{\pi}{2} \quad\Rightarrow\quad x \,=\,\tfrac{\pi}{10}


    I know . . . I oversimplified the problem.
    . . But you can generalize the solution, right?

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