Can anyone please explain how I can solve the following problem:

Solve for \(x\):

\[ \cos(\cos(\cos(\cos x))) = \sin(\sin(\sin(\sin x))) \]

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## Comments

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TopNewestWell, according to Wolfram alpha, the solution(s) are:

\(\begin{eqnarray} x & \approx & -4.71 \pm 1.21 i \\ x & \approx & 0.76 \pm 0.61i \end{eqnarray}\)

Its graph looks like this:

I couldn't properly solve the problem, but I managed to get to this point.

\(\cos(\cos(\cos(\cos x))) = \sin(\sin(\sin(\sin x))) \\ \implies \sqrt{1- \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}}} = \sqrt{1- \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}}} \\ \implies \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} \\ \implies \cos^2 \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sin^2 \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = 0 \\ \implies \bigg[ \cos \left( \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} + \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} \right) \bigg] \cdot \bigg[ \cos \left( \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} \right) \bigg] = 0 \qquad \qquad \small \color{blue}{\text{Using} \ \cos^2 A - \sin^2 B = \cos(A+B) \cdot \cos(A-B)}\)

Therefore, the trivial solutions are of form:

\(1. \ \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} + \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \dfrac{(2n+1) \pi}{2} \qquad \small \color{blue}{\text{Where} \ n \in \mathbb{N}}\)

\(2. \ \sqrt{1- \cos^2 \sqrt {1- \cos^2 \sqrt{1- \cos^2 x}}} - \sqrt{1- \sin^2 \sqrt {1- \sin^2 \sqrt{1- \sin^2 x}}} = \dfrac{(2m+1) \pi}{2} \qquad \small \color{blue}{\text{Where} \ m \in \mathbb{N}}\)

The two pairs of answers which are complex will be of the forms as discussed above.

I don't know of a better way to approach this problem.

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Is there any other way? Perhaps, since we know that sin(x) = cos(x) for x = \frac{{pi}{4} and \frac{5 * {pi}}{4}. Or that sin({pi} - x) = cos(x). Do we get any lead like that? I couldn't find any, but you might. Thanks for helping.

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Well, you might get principal solutions like that. But you have to think for general cases while solving problems like these, for example \(\sin x = \cos x\) has the general solution of \(n \pi + \dfrac{\pi}{4}\), where \(n \in \mathbb{N}\). As for the identity \(\sin^k \left( \dfrac{\pi}{2} - x \right) = \cos^k x\), I haven't still tried that approach. I would check it out too, and let you know.

I found it online that this is a problem from the

All Russian Olympiad 1995, Grade 10, Problem 1.Log in to reply

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What have you done? What have you tried? Where are you stuck?

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This is a problem which apparently came in one of Russia's elite math examinations (I have no idea which one) Source: The Art And Craft Of Problem Solving

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However, I graphed these two functions and found that they never intersect, implying there are no solutions. But, how do we theoretically prove that this has no solutions within the real numbers?

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