How do we know that is not possible to invert $x=t+\cos t$ analytically? From this answer to how to get a get a nice “cosine looking” curve following the y=x direction?

It is not possible to invert $x=t+\cos t$ analytically.

I'm sure it's true and I wouldn't know how to try, but how can the impossibility be shown?
 A: 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
A: 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.
A: It can be shown by simple rearranging the equation, that the equation is not in a form that is solvable in terms of Lambert W.
The equation is an equation of elementary functions. You are looking for the inverse (means the inverse function) of the elementary function $t\mapsto t+\cos(t)$ over domains where the function is bijective.
The main theorem in [Ritt 1925], that's proven also in [Risch 1979], lets suggest that the elementary function above cannot have an elementary inverse. The main theorems in [Lin 1983] and [Chow 1999] imply, assuming Schanuel's conjecture is true, that your equation cannot have solutions that are elementary numbers or explicit elementary numbers if $x$ is an elementary number. This implies, assuming Schanuel's conjecture is true, that the elementary function above over non-discrete domains cannot have an elementary inverse therefore.
$\ $
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
