How do we know that is not possible to invert $x=t+\cos t$ analytically?

Solution 1:

In this case it can be shown by the inverse function theorem. Since $(t+\cos t)' = 1-\sin t$ is zero at $t=2\pi n + \pi/2$ with $n \in \mathbb{Z}$, in a neighborhood of these points the function is not analytically invertible. These are in fact the points of vertical tangent (infinite slope) in the graph from the linked answer.

Solution 2:

For the purpose of the question, inversion of $x = t + \cos t$ analytically is formally defined as whether or not there exists a representation, in terms of elementary operations, of $t$ purely as a function of $x$.

The space of mathematical expressions can be represented as a syntax tree. As well, algebraic equivalences can be represented as modifications on that syntax tree. The problem then becomes, given this set of possible transitions and replacements, is it possible to arrive at a syntax tree which obeys this set of rules, in this case, a syntax tree where the LHS is $t$ and the RHS does not contain $x$. Proving this can likely be done using graph theory, and is the general approach of Mathematica, which does imply, but does not assert, that this representation does not exist