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Math Help - Solving for Interval [0, 2pi)

  1. #1
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    Trig Product Formula Problem

    My question states: cos(3x)+cos(x)=0
    I got two answer pi/2 and pi/4 but there are supposed to be more, can you show me how to get all the answers?
    Thanks alot
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  2. #2
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    Solving for Interval [0, 2pi)

    This problem was pretty tough for me:

    cos(2x)+2cos(x)=0

    can you show me how to do this? Thanks!

    -Aussiekid90
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by aussiekid90
    This problem was pretty tough for me:

    cos(2x)+2cos(x)=0

    can you show me how to do this? Thanks!

    -Aussiekid90
    Recall that cos(2x)=2cos^2(x)-1

    Thus
    cos(2x)+2cos(x)=0
    2cos^2(x)-1+2cos(x)=0
    2cos^2(x)+2cos(x)-1=0

    This is a quadratic. Let y = cos(x):
    2y^2+2y-1=0

    Using the quadratic formula I get:
    y = \frac{-2 \pm \sqrt{2^2-4*2*-1}}{2*2}
    y = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2}

    Thus cos(x) = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}

    Now. -\frac{1}{2} - \frac{\sqrt{3}}{2} < -1 so we need to take the positive sign. As it happens the result doesn't seem to have a nice form, so:

    x = cos^{-1} \left ( -\frac{1}{2} + \frac{\sqrt{3}}{2} \right ) \approx 1.19606 \, rad. Since this angle is second quadrant, we also get an angle x = 2 \pi \, rad - 1.19606 \, rad = 5.08712 \, rad.

    -Dan
    Last edited by topsquark; May 8th 2006 at 01:35 PM. Reason: LaTeX typo!
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  4. #4
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    Quote Originally Posted by aussiekid90
    My question states: cos(3x)+cos(x)=0
    I got two answer pi/2 and pi/4 but there are supposed to be more, can you show me how to get all the answers?
    Thanks alot
    Use identity,
    \cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)
    Thus,
    \cos 3x+\cos x=0
    Thus,
    2\cos 2x\cos x=0
    Thus,
    \cos 2x\cos x=0
    Thus, set factors equal to zero.
    \left\{ \begin{array}{c} \cos 2x=0\\ \cos x=0
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  5. #5
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    You already asked that question in "Urget Math". DO NOT MAKE DOUBLE POSTS
    -=USER WARNED=-
    The thread has been merged.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker
    You already asked that question in "Urget Math". DO NOT MAKE DOUBLE POSTS
    -=USER WARNED=-
    The thread has been merged.
    One problem is:
    cos(2x)+2cos(x)=0

    and the other is:
    cos(3x)+cos(x)=0

    They're close, but not quite the same.

    -Dan
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