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  1. #1
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    solve

    cos x cos 2x cos 3x=1
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by perash View Post
    cos x cos 2x cos 3x=1
    The only way I can construct a solution is if
    cos(x) = 1
    and
    cos(2x) = 1
    and
    cos(3x) = 1

    or any two of these is equal to -1 and the remaining is equal to 1.

    The solution to the first set of equations is, for example
    x = \frac{\pi}{2} + 2n \pi
    and
    x = \frac{\pi}{4} + k \pi
    and
    x = \frac{\pi}{6} + p \frac{2\pi}{3}
    respectively where n, k, and p are integers.

    So:
    k = 2n + \frac{1}{4}
    and
    p = 3n + \frac{1}{2}

    Since k and p have to be integral there can be no solutions of this kind.

    See what the other two solution options give you.

    -Dan
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  3. #3
    Senior Member JaneBennet's Avatar
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    Let \mathrm{f}(x)=\cos{x}\cos{2x}\cos{3x}. We need only consider 0\leq x < \pi since \mathrm{f}(x) has period \pi (i.e. \mathrm{f}(x+\pi)=\mathrm{f}(x)).

    For x\in[0,\pi), \cos{x} is equal to ±1 only when x=0. And indeed \mathrm{f}(0)=1. So that’s the only solution in that interval, and since f has period \pi, the complete solution set is \{n\pi:n\in\mathbb{Z}\}.
    Last edited by JaneBennet; March 17th 2008 at 04:29 AM.
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