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There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of $f$. And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$. Then $f(f(x))$ is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not.

As for discontinuous ones, there are terribly many of them ($2^{\mathbb R}$) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$, but I doubt this would count as a solution of the equation.

UPDATE: Here is how to construct a solution (this is technical and I might overlook something).

Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$. Define the following equivalence relation on $X$: $x$ and $y$ are equivalent iff $g^n(x)=g^m(y)$ for some positive integers $m$ and $n$. Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with $g$. If $Y$ and $Z$ are two similar orbits, one can define $f$ on $Y\cup Z$ as follows: on $Y$, $f$ is that bijection to $Z$, and on $Z$, $f$ is the inverse bijection composed with $g$.

So if the orbits can be split into pairs of similar ones, we have a desired $f$. Now remove from the real line the fixed point of cos and all its roots ($\pi/2$ and the like). Then, if I am not missing something, in the remaining set $X$ all orbits of $\cos$ are similar, so we can define $f$ as above. Define $f$ so that $0$ has a nonempty pre-image (that is, the orbit containing $0$ should be used as $Z$ and not as $Y$). Finally, map the fixed point of $\cos$ to itself, and the roots of $\cos$ to some pre-image of $0$.

There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of $f$. And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$. Then $f(f(x))$ is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not.

As for discontinuous ones, there are terribly many of them ($2^{\mathbb R}$) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$, but I doubt this would count as a solution of the equation.

UPDATE: Here is how to construct a solution (this is technical and I might overlook something).

Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$. Define the following equivalence relation on $X$: $x$ and $y$ are equivalent iff $g^n(x)=g^m(y)$ for some positive integers $m$ and $n$. Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with $g$. If $Y$ and $Z$ are two similar orbits, one can define $f$ on $Y\cup Z$ as follows: on $Y$, $f$ is that bijection to $Z$, and on $Z$, $f$ is the inverse bijection composed with $g$.

So if the orbits can be split into pairs of similar ones, we have a desired $f$. Now remove from the real line the fixed point of $\cos$ and all its roots ($\pi/2$ and the like). Then, if I am not missing something, in the remaining set $X$ all orbits of $\cos$ are similar, so we can define $f$ as above. Define $f$ so that $0$ has a nonempty pre-image (that is, the orbit containing $0$ should be used as $Z$ and not as $Y$). Finally, map the fixed point of $\cos$ to itself, and the roots of $\cos$ to some pre-image of $0$.

EDIT (by BP July 17, 2023): When removing the fixed point $t$ of $\cos$ above, one also needs to remove the grand orbit of $t$ under $\cos$ and define $f$ separately on this grand orbit, which is $\pm t + \mathbb{Z} \pi$. One possibility is to set $$f(\pm t + n\pi) = \begin{cases} t, &\text{ if $n=0$;}\\ -t, &\text{ if $n$ is even and nonzero; and}\\ t+2\pi, &\text{ if $n$ is odd.} \end{cases}$$

There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of f. And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$. Then f(f(x)) is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not.

As for discontinuous ones, there are terribly many of them ($2^{\mathbb R}$) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$, but I doubt this would count as a solution of the equation.

UPDATE: Here is how to construct a solution (this is technical and I might overlook something).

Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$. Define the following equivalence relation on X: x and y are equivalent iff $g^n(x)=g^m(y)$ for some positive integers m and n. Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with g. If Y and Z are two similar orbits, one can define f on $Y\cup Z$ as follows: on Y, f is that bijection to Z, and on Z, f is the inverse bijection composed with g.

So if the orbits can be split into pairs of similar ones, we have a desired f. Now remove from the real line the fixed point of cos and all its roots ($\pi/2$ and the like). Then, if I am not missing something, in the remaining set X all orbits of cos are similar, so we can define f as above. Define f so that 0 has a nonempty pre-image (that is, the orbit containing 0 should be used as Z and not as Y). Finally. map the fixed point of cos to itself, and the roots of cos to some pre-image of 0.

There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of $f$. And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$. Then $f(f(x))$ is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not.

As for discontinuous ones, there are terribly many of them ($2^{\mathbb R}$) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$, but I doubt this would count as a solution of the equation.

UPDATE: Here is how to construct a solution (this is technical and I might overlook something).

Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$. Define the following equivalence relation on $X$: $x$ and $y$ are equivalent iff $g^n(x)=g^m(y)$ for some positive integers $m$ and $n$. Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with $g$. If $Y$ and $Z$ are two similar orbits, one can define $f$ on $Y\cup Z$ as follows: on $Y$, $f$ is that bijection to $Z$, and on $Z$, $f$ is the inverse bijection composed with $g$.

So if the orbits can be split into pairs of similar ones, we have a desired $f$. Now remove from the real line the fixed point of cos and all its roots ($\pi/2$ and the like). Then, if I am not missing something, in the remaining set $X$ all orbits of $\cos$ are similar, so we can define $f$ as above. Define $f$ so that $0$ has a nonempty pre-image (that is, the orbit containing $0$ should be used as $Z$ and not as $Y$). Finally, map the fixed point of $\cos$ to itself, and the roots of $\cos$ to some pre-image of $0$.

There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of f. And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$. Then f(f(x)) is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not.

As for discontinuous ones, there are terribly many of them ($2^{\mathbb R}$) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$, but I doubt this would count as a solution of the equation.

UPDATE: Here is how to construct a solution (this is technical and I might overlook something).

Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$. Define the following equivalence relation on X: x and y are equivalent iff $g^n(x)=g^m(y)$ for some positive integers m and n. Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with f. If Y and Z are two similar orbits, one can define f on $Y\cup Z$ as follows: on Y, f is that bijection to Z, and on Z, f is the inverse bijection composed with f.

So if the orbits can be split into pairs of similar ones, we have a desired f. Now remove from the real line the fixed point of cos and all its roots ($\pi/2$ and the like). Then, if I am not missing something, in the remaining set X all orbits are similar, so we can define f as above. Define f so that 0 has a nonempty pre-image (that is, the orbit containing 0 should be used as Z and not as Y). Finally. map the fixed point of cos to itself, and the roots of cos to some pre-image of 0.

There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$), it should be a fixed point of f. And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$. Then f(f(x)) is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not.

As for discontinuous ones, there are terribly many of them ($2^{\mathbb R}$) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$, but I doubt this would count as a solution of the equation.

UPDATE: Here is how to construct a solution (this is technical and I might overlook something).

Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$. Define the following equivalence relation on X: x and y are equivalent iff $g^n(x)=g^m(y)$ for some positive integers m and n. Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with g. If Y and Z are two similar orbits, one can define f on $Y\cup Z$ as follows: on Y, f is that bijection to Z, and on Z, f is the inverse bijection composed with g.

So if the orbits can be split into pairs of similar ones, we have a desired f. Now remove from the real line the fixed point of cos and all its roots ($\pi/2$ and the like). Then, if I am not missing something, in the remaining set X all orbits of cos are similar, so we can define f as above. Define f so that 0 has a nonempty pre-image (that is, the orbit containing 0 should be used as Z and not as Y). Finally. map the fixed point of cos to itself, and the roots of cos to some pre-image of 0.