What is the solution of $\cos(x)=x$?

The equation in question is a transcendental equation. Apart of guessing, numerical or analytical methods, there is no way of solving the equation without using another transcendental function, and therefore argue in circles.

In this case, denote $g(x)=\cos x -x$, see that its derivative is negative with countable many zeros, and therefore $g$ is strictly decreasing, yielding that there is at most one solution to $g(x)=0$. Since $g(0)g(\pi/2)<0$ there is such a solution. Arbitrary precise approximations can be found using Newton, bisection, or false position method.

As user Myself commented, it is a challenge (not so hard) to prove that the sequence $x_{n+1}=\cos x_n, x_0 \in \Bbb{R}$ converges to the unique solution to $\cos x=x$.

Another related problem which I encountered last week when trying to help one of my friends for an exam is to find all continuous functions $f : \Bbb{R} \to \Bbb{R}$ with the property that $f(x)=f(\cos x)\ \forall x \in \Bbb{R}$.

Mathworld calls this the Dottie Number. The page makes no mention of existence/non-existence of "closed" form and I would guess it is still open.

Remembering the Kepler equation and its solution, the Dottie number can be analytically written as:

$$D = 2\sum_{n=0}^\infty \left( \frac{J_{4n+1}(4n+1)}{4n+1} - \frac{J_{4n+3}(4n+3)}{4n+3}\right)$$

where $J_{n}$ are the Bessel functions. Such series is convergent and can be evaluated numerically.

A proof and numerical evaluations are provided in :

Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$