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

Solve for \(x\):

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

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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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. – Tapas Mazumdar · 8 months ago

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– Nilabha Saha · 8 months ago

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.Log in to reply

I found it online that this is a problem from the

All Russian Olympiad 1995, Grade 10, Problem 1. – Tapas Mazumdar · 8 months agoLog in to reply

– Nilabha Saha · 8 months ago

Yes, I found the with the citation of 'Russia' in the book 'The Art And Craft Of Problem Solving'.Log in to reply

What have you done? What have you tried? Where are you stuck? – Calvin Lin Staff · 8 months ago

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– Nilabha Saha · 8 months ago

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 SolvingLog in to reply

– Nilabha Saha · 8 months ago

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?Log in to reply

– Calvin Lin Staff · 8 months ago

Yup, the idea is to show that these 2 graphs never intersect (over the reals). It's somewhat tricky to prove that though. AFAIK there isn't a "non-calculus" approach.Log in to reply

– Nilabha Saha · 8 months ago

Can you guide me on the calculus approach?Log in to reply