Solving $ \cos (\cos (\cos (\cos(x))))=\sin (\sin (\sin (\sin (x)))) $ I found this problem in a collection of contest problems of a Russian competition in 1995 and wasn't able to solve it.

Solve for real $x$:
  $$ \cos (\cos (\cos (\cos(x))))=\sin (\sin (\sin (\sin (x)))) $$

My guess is that there is no solution, but how do I prove it? I tried to estimate
$LHS\ge \cos (1) \ge \cos(\pi/3)=1/2 $
and RHS similarly but the ranges overlap..
Do you have a better idea?
 A: This is not a solution, but here's a pretty compelling picture of the the fourth iterates of cosine and sine (in blue and red, respectively).
It suggests that you can't uniformly bound the two apart from one another.  (The functions appear to have slightly overlapping ranges).

A: Let $f(x)=\cos(\cos(\cos(\cos x)))-\sin(\sin(\sin(\sin x)))$
For $x\in[\pi,2\pi], f(x)>0$ As cos term positive and sin term negative !
Also $f(x)=f(\pi-x)\implies \text{it is sufficient to solve it for}\,\,x\in[0,\pi/2]$
$\implies \cos x+ \sin x\le 
\sqrt 2<\frac{\pi}{2}\,\,\,\text{  or }\cos x<\frac{\pi}{2}–\sin x$
$\implies \cos(\cos x)>\sin(\sin x)\implies \cos(\cos(\cos x))<\cos(\sin(\sin x))$
$\implies \cos(\cos(\cos x))+\sin(\sin(\sin x))<\cos(\sin(\sin x)) +\sin(\sin(\sin x)) \le \sqrt2<\frac{\pi}{2}$
$\implies \cos(\cos(\cos x))<\frac{\pi}{2}-\sin(\sin(\sin x))$
$\implies\cos(\cos(\cos(\cos x)))>\sin(\sin(\sin(\sin x)))$
Hence no solution in real numbers !
A: Rephrase the problem as:
$$ \sin( \pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) ) = \sin(\sin(\sin(\sin(x))))$$
Our strategy will be to assume the expression above is true for some $x$, "invert" the sine on both sides, and show that the resulting expression cannot possibly have a solution. Doing this process once, we have
\begin{align}
\pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) & = \sin(\sin(\sin(x))) + 2\pi n \\
\textrm{or} & \\
\pi/2 + \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) & = \pi - \sin(\sin(\sin(x))) + 2\pi n
\end{align}
for some integer $n$, which implies
$$ \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) =  \pm (\sin(\sin(\sin(x))) + 2 \pi (n - 1/4)),$$
where the $\pm$ covers the two cases above, and indicate that at least one of the two possibilities must hold, but not necessarily both. We must have $n = 0$ since both sine terms must have values in $[-1,1]$. So
$$ \sin(\pi/2 + \sin(\pi/2 + \sin(\pi/2 + x) ) ) =  \pm (\sin(\sin(\sin(x))) - \pi/2),$$
and so it must be that both sides of the above expression lie in either the interval $[-1, 1 - \pi/2]$ or the interval $[-1 + \pi/2, 1]$. Applying our strategy once more, we have
$$\sin(\pi/2 + \sin(\pi/2 + x) ) = \pm (\arcsin( \sin(\sin(\sin(x))) - \pi/2) - \pi/2).$$
On the left side, clearly the expression must be in $[-1,1]$, while on the right side, the expression must be either in the interval $[-\pi, \arcsin(1 - \pi/2) - \pi/2]$ or in the interval $[\pi/2 - \arcsin(1 - \pi/2), \pi]$.
We derive a contradiction by showing that neither of these two intervals intersect with $[-1,1]$. In particular, we show that $\arcsin(1 - \pi/2) - \pi/2 < -1$. Using the fact that $\pi > 3$,
$$ \arcsin(1 - \pi/2) - \pi/2 < \arcsin(1 - 3/2) - \pi/2 = -\pi/6 - \pi/2 = - 2\pi/3 < -1.$$
