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Math Help - Help me to solve the trigonometric equation

  1. #1
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    Question Help me to solve the trigonometric equation

    Hi I have tried to solve this... (but my teacher never gets satisfied).

    Solve the trigonometric equation:
    4sinx*sin2x*sin4x=sin3x

    My solution so far...

    4sinx*sin2x*sin4x−sin3x=0

    3x=sinx(3-4((sinx)^2)
    0=4sinx*sin2x*sin4x-sinx(3-4*((sinx)^2)= sinx(4sin(2x)) *sin4x-(3-4*((sinx)^2)

    So:
    0=sinx

    or
    0=4sin2x*sin4x-(3-4*((sinx)^2)

    Please help.
    Last edited by sf1903; January 10th 2011 at 09:54 AM.
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  2. #2
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    Quote Originally Posted by sf1903 View Post
    Solve the trigonometric equation:
    4sinx*sin2x*sin4x=sin3x
    Can I suggest

    \displaystyle \sin 2x = 2\sin x \cos x

    \displaystyle \sin 3x  = 3\sin x -4 \sin^3 x

    \displaystyle \sin 4x = 2\sin 2x \cos 2x = 2(2\sin x \cos x )(1-2\sin^2 x)
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Can I suggest

    \displaystyle \sin 2x = 2\sin x \cos x

    \displaystyle \sin 3x  = 3\sin x -4 \sin^3 x

    \displaystyle \sin 4x = 2\sin 2x \cos 2x = 2(2\sin x \cos x )(1-2\sin^2 x)

    Should I then insert it to the original equation? 4sinx*sin2x*sin4x=sin3x

    So:
    4*Sinx(2SinxCosx)*2(2SinxCosx)(1-2Sin^2x)=3Sinx-4Sin^3x

    4Sinx*3(2Sinx*Cosx)(1-2*Sin^2x)=3Sinx-4Sin^3x

    4Sinx*3(2SinxCosx)(1-2Sin^2x)-3Sinx+4Sin^3x=0

    ?
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  4. #4
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    Quote Originally Posted by sf1903 View Post
    Hi I have tried to solve this... (but my teacher never gets satisfied).

    Solve the trigonometric equation:
    4sinx*sin2x*sin4x=sin3x

    My solution so far...

    4\sin x\sin2x\sin4x-\sin3x=0

    \sin3x=\sin x(3-4\sin^2x)

    0=4\sin x\sin2x\sin4x-\sin x(3-4\sin^2x)= \sin x(4\sin2x\sin4x-(3-4\sin^2x))

    So:

    0=\sin x
    or
    0=4\sin2x\sin4x-(3-4\sin^2x)
    Good start! The first alternative, 0=\sin x, is easy to deal with, so let's look at the second one, 0 = 4\sin2x\sin4x - (3-4\sin^2x).

    Use the trig identity \cos(\theta-\phi) + \cos(\theta+\phi) = 2\sin\theta\sin\phi. It tells you that 2\sin2x\sin4x = \cos2x - \cos6x, and 2\sin^2x = 1-\cos2x. Therefore

    0 = 4\sin2x\sin4x - (3-4\sin^2x) = 2(\cos2x - \cos6x) - (3 - 2(1-\cos2x)) = -2\cos6x-1.

    Now finish it off from there.
    Last edited by Opalg; January 13th 2011 at 07:16 AM. Reason: missing parenthesis
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  5. #5
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    Quote Originally Posted by sf1903 View Post
    Should I then insert it to the original equation? 4sinx*sin2x*sin4x=sin3x

    So:
    4*Sinx(2SinxCosx)*2(2SinxCosx)(1-2Sin^2x)=3Sinx-4Sin^3x

    4Sinx*3(2Sinx*Cosx)(1-2*Sin^2x)=3Sinx-4Sin^3x

    4Sinx*3(2SinxCosx)(1-2Sin^2x)-3Sinx+4Sin^3x=0

    ?

    This expands further...
    Last edited by pickslides; January 11th 2011 at 11:45 AM. Reason: *sp
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  6. #6
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    Quote Originally Posted by Opalg View Post
    Good start! The first alternative, 0=\sin x, is easy to deal with, so let's look at the second one, 0 = 4\sin2x\sin4x - (3-4\sin^2x).

    Use the trig identity \cos(\theta-\phi) + \cos(\theta+\phi) = 2\sin\theta\sin\phi. It tells you that 2\sin2x\sin4x = \cos2x - \cos6x, and 2\sin^2x = 1-\cos2x. Therefore

    0 = 4\sin2x\sin4x - (3-4\sin^2x) = 2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.

    Now finish it off from there.
    So:
    2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.

    Therefore:
    2(\cos2x - \cos6x) - (3 - 2(1-\cos2x)+2\cos6x-1=0<br />

    How should I continue?
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  7. #7
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    Quote Originally Posted by sf1903 View Post
    So:
    2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.

    Therefore:
    2(\cos2x - \cos6x) - (3 - 2(1-\cos2x)+2\cos6x-1=0<br />

    How should I continue?
    What's with the ? You have been told exactly how to do it!:

    Quote Originally Posted by opalg
    0 = 4\sin2x\sin4x - (3-4\sin^2x) = 2(\cos2x - \cos6x) - (3 - 2(1-\cos2x) = -2\cos6x-1.
    So the equation you have to solve is obviously 0 = -2 \cos(6x) - 1. Surely you can solve this ....
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