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Math Help - general solution qs

  1. #1
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    general solution qs

    write general solution of

    sin4x - sin2x = 0
    cos3x = cos x
    sin3x = sin2x

    ack, i dont realli no wat sin4x or cos3x or sin3x is, or how to get to it, so if anybody could help me, i would greatly appreciate it

    thanx !
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  2. #2
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    Quote Originally Posted by iiharthero View Post
    write general solution of

    sin4x - sin2x = 0
    cos3x = cos x
    sin3x = sin2x

    ack, i dont realli no wat sin4x or cos3x or sin3x is, or how to get to it, so if anybody could help me, i would greatly appreciate it

    thanx !
    Use the following:

    \sin A = \sin B \Rightarrow A = B + 2 n \pi or A = (\pi - B) + 2n \pi where n is an integer.

    \cos A = \cos B \Rightarrow A = B + 2 n \pi or A = - B + 2n \pi where n is an integer.
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by iiharthero View Post
    write general solution of

    sin4x - sin2x = 0
    sin4x-sin2x=0

    2sin2xcos2x-sin2x=0

    sin2x(2cos2x-1)=0

    either sin2x=0.........(1)

    or cos2x=\frac{1}{2}..............(2)

    for 1st case,
    sin2x=0

    2sinxcosx=0

    for 2nd case,
    cos2x=\frac{1}{2}

    2cos^2x-1=\frac{1}{2}
    Last edited by mr fantastic; June 20th 2009 at 03:24 PM. Reason: Fixed a latex tag
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  4. #4
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    Quote Originally Posted by iiharthero View Post
    cos3x = cos x
    There is a little-known formula for cos 3x:
    \cos 3x = 4\cos^3 x - 3\cos x

    Here is how it's derived:
    \begin{aligned}<br />
\cos 3x &= \cos (2x + x) \\<br />
&= \cos 2x \cos x - \sin 2x \sin x \\<br />
&= (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x \\<br />
&= 2\cos^3 x - \cos x - 2\cos x \sin^2 x \\<br />
&= 2\cos^3 x - \cos x - 2\cos x(1 - \cos^2 x) \\<br />
&= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x \\<br />
&= 4\cos^3 x - 3\cos x<br />
\end{aligned}

    Back to the problem:
    \begin{aligned}<br />
\cos 3x &= \cos x \\<br />
4\cos^3 x - 3\cos x &= \cos x \\<br />
4\cos^3 x - 4\cos x &= 0 \\<br />
4\cos x(\cos^2 x - 1) &= 0 \\<br />
 4\cos x(\cos x + 1)(\cos x - 1) &= 0<br />
 \end{aligned}

    Now set each factor equal to 0 and solve for x...


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  5. #5
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    Quote Originally Posted by yeongil View Post
    There is a little-known formula for cos 3x:
    \cos 3x = 4\cos^3 x - 3\cos x

    Here is how it's derived:
    \begin{aligned}<br />
\cos 3x &= \cos (2x + x) \\<br />
&= \cos 2x \cos x - \sin 2x \sin x \\<br />
&= (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x \\<br />
&= 2\cos^3 x - \cos x - 2\cos x \sin^2 x \\<br />
&= 2\cos^3 x - \cos x - 2\cos x(1 - \cos^2 x) \\<br />
&= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x \\<br />
&= 4\cos^3 x - 3\cos x<br />
\end{aligned}

    Back to the problem:
    \begin{aligned}<br />
\cos 3x &= \cos x \\<br />
4\cos^3 x - 3\cos x &= \cos x \\<br />
4\cos^3 x - 4\cos x &= 0 \\<br />
4\cos x(\cos^2 x - 1) &= 0 \\<br />
4\cos x(\cos x + 1)(\cos x - 1) &= 0<br />
\end{aligned}

    Now set each factor equal to 0 and solve for x...


    01
    The style of question and the fact that the OP states: "ack, i dont realli (sic) no (sic) wat (sic) sin4x or cos3x or sin3x is"

    suggests that the approach advised in my first post is probably the best one for the OP to use.
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  6. #6
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    Exactly. If is not necessary to change to "sin(x)" and "cos(x)". To solve "sin(3x)= sin(2x)", just write 3x= 2x+ 2k\pi.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Exactly. If is not necessary to change to "sin(x)" and "cos(x)". To solve "sin(3x)= sin(2x)", just write 3x= 2x+ 2k\pi
    and 3x = (\pi - 2x) + 2k \pi

    where k is an integer.
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