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    Well, if we factorize the LHS, we can further restrict the domain to (\frac{\pi}{3},\frac{\pi}{2}] but I don't see how that helps...

    Do you have a context for this question, Plato's Trousers?
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    While I've got this equation written down (again) I'll post it. I'm not sure it's much help but it's somewhere safe to write it down.

    32c^5-32c^3+3c+1=0

    and the factorisation

    (4c^2-2c-1)(8c^3+4c^2-4c-1)=0
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    (Original post by BabyMaths)
    While I've got this equation written down (again) I'll post it. I'm not sure it's much help but it's somewhere safe to write it down.

    32c^5-32c^3+3c+1=0

    and the factorisation

    (4c^2-2c-1)(8c^3+4c^2-4c-1)=0
    The factorisation is not right. Expanded you get - 32c^5-32c^3+6c+1=0
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    (Original post by SecondHand)
    The factorisation is not right. Expanded you get - 32c^5-32c^3+6c+1=0
    The factorisation is right. I just wrote the wrong equation.
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    Yeah, I got the same equation (all signs flipped).

    The determinant of the quadratic is negative, right? So the quadratic has no real solutions.

    Then we can...use the cubic formula and find the real solution(s) to c, and cancel out or double the solutions when we put c=cos(±x+2pik) into the domain [0,pi/2].

    Sum-to-product, quintic factorization, and cubic formula...I'm still waiting for the context that might lead us to a more elegant solution. :P
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    (Original post by aznkid66)
    Yeah, I got the same equation (all signs flipped).

    The determinant of the quadratic is negative, right? So the quadratic has no real solutions.

    Then we can...use the cubic formula and find the real solution(s) to c, and cancel out or double the solutions when we put c=cos(±x+2pik) into the domain [0,pi/2].

    Sum-to-product, quintic factorization, and cubic formula...I'm still waiting for the context that might lead us to a more elegant solution. :P
    The determinant is positive, 4-(4*4*-1)=8
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    (Original post by SecondHand)
    The determinant is positive, 4-(4*4*-1)=8
    D'oh. I always mess up when I take the negative of a negative.

    So cosx=(1/4)±(1/sqrt(8)) ...that looks nice :|
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    (Original post by aznkid66)
    D'oh. I always mess up when I take the negative of a negative.

    So cosx=(1/4)±(1/sqrt(8)) ...that looks nice :|
    You just need to show that only two solutions exist, not find the solutions.
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    (Original post by SecondHand)
    You just need to show that only two solutions exist, not find the solutions.
    Well, cubic aside, even after finding there exists two unique solutions for c we don't know how many unique x values of each solution c=cosx are in the domain of 0≤x≤pi/2.
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    Could you not just say, within the interval, there is only one point of inflection (and then find that point which is blow -1/2), hence there can only be two points since you've already proved the graph is continuous.
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    (Original post by djpailo)
    Could you not just say, within the interval, there is only one point of inflection (and then find that point which is blow -1/2), hence there can only be two points since you've already proved the graph is continuous.
    But how do you prove that there is only one through graphical analysis?
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    (Original post by djpailo)
    Could you not just say, within the interval, there is only one point of inflection (and then find that point which is blow -1/2), hence there can only be two points since you've already proved the graph is continuous.
    Yes. This is the method they are looking for.
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    (Original post by Plato's Trousers)
    Yes. This is the method they are looking for.
    So, without calculus?
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    (Original post by ghostwalker)
    So, without calculus?
    yes, I think it's basically about finding values of the function either side of zero and then inferring that (because the function is continuous) that it must have zeroes between them
Updated: June 30, 2012
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