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Math Help - Solve cos^2(...)+cos^2(...)-2 = 0 ?

  1. #1
    Junior Member pinion's Avatar
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    Solve cos^2(...)+cos^2(...)-2 = 0 ?

    anyone have any clever ideas on how to solve this for x?

    cos^2(\pi*16^x)+cos^2(\pi*16^{1-x})-2 = 0

    ?
    (cos(\pi*16^x)-1)(cos(\pi*16^x)+1)=-(cos(\pi*16^{1-x})-1)(cos(\pi*16^{1-x})+1)
    ?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by pinion View Post
    anyone have any clever ideas on how to solve this for x?

    cos^2(\pi*16^x)+cos^2(\pi*16^{1-x})-2 = 0

    ?
    (cos(\pi*16^x)-1)(cos(\pi*16^x)+1)=-(cos(\pi*16^{1-x})-1)(cos(\pi*16^{1-x})+1)
    ?
    Since |\cos(x)| \le 1 this equation implies that:

    \cos^2(\pi*16^x)=1

    and

    \cos^2(\pi*16^{1-x})=1

    RonL
    Last edited by CaptainBlack; February 10th 2008 at 10:44 PM.
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  3. #3
    Junior Member pinion's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Since |\cos(x)| \le 1 this equation implies that:

    \cos^2(\pi*16^x)=1

    and

    \cos^2(\pi*16^{1-x})=1

    RonL
    huh?
    Last edited by CaptainBlack; February 10th 2008 at 10:44 PM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by pinion View Post
    huh?
    Your equation asserts that the sum of two positive quataties that are both
    less than or equal to 1 is 2, so both must be exactly 1.

    RonL

    PS huh? is uninformative, please explain yourself more clearly in future
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    Since |\cos(x)| \le 1 this equation implies that:

    \cos^2(\pi*16^x)=1

    and

    \cos^2(\pi*16^{1-x})=1

    RonL
    If you require further help in finding the solutions let us know, but the problem is trivial from here.

    RonL
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  6. #6
    Junior Member pinion's Avatar
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    Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know. But I do know this follows from that observation -
    cos(\pi*16^x) = 1
    and
    cos(\pi*16^x) = -1
    and
    cos(\pi*16^{1-x}) = 1
    and
    cos(\pi*16^{1-x}) = -1
    and since the arccos of both 1 and -1 is c*\pi (or more specifically c_1*\pi and c_2*\pi where c_1 is even and c_2 is odd) I would now have to find c_1,c_2,c_3,c_4 - so unless there is an algebraic method for doing that, splitting the equation in two doesn't help me. Is there anything else you can recommend?
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    Grand Panjandrum
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    Quote Originally Posted by pinion View Post
    Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know. But I do know this follows from that observation -
    cos(\pi*16^x) = 1
    and
    cos(\pi*16^x) = -1
    and
    cos(\pi*16^{1-x}) = 1
    and
    cos(\pi*16^{1-x}) = -1
    and since the arccos of both 1 and -1 is c*\pi (or more specifically c_1*\pi and c_2*\pi where c_1 is even and c_2 is odd) I would now have to find c_1,c_2,c_3,c_4 - so unless there is an algebraic method for doing that, splitting the equation in two doesn't help me. Is there anything else you can recommend?
    That:

    <br />
\cos^2(\pi*16^x)=1<br />

    implies that \pi 16^x=k \pi for some integer k, which then implies that 16^x=2^{4x}=k but this implies that:

    4x=\log_2(k)/2

    or:

    x=\log_2(k)/8.

    Similarly (using the other equation): we find that for some integer n:

    1-x=\log_2(n)/8.

    So:

    \log_2(k)/8=1-\log_2(n)/8

    or:

    \log_2(k/n)=8

    so:

    k/n=2^8.

    so every integer n determines a solution x=1-\log_2(n)/8.

    (don't rely on the algebra above being right as I cannot check it in detail at present, either I or another helper will correct any mistakes when we get the chance).

    RonL
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by pinion View Post
    Thanks for being so nice about me not explaining in the manner you require, my bad. But as to the 'huh', I was firstly confused at how your statement followed from the equation, and secondly confused at why I would/should observe that fact to solve the equation. Is that the only way? Or do you suggest that to be the easiest way? I don't know.
    The reason why the obsevation is worth making is that it is useful information
    about the nature of the solutions that is otherwise not obvious.

    Other approaches are likely to lead you through some horendous trig transformations that are unlikely to go anywhere.

    RonL
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  9. #9
    Junior Member pinion's Avatar
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    thanks, i'll look into trig transformations.
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