About the algebraic character of the solution of $\cos x=x$

Question

The equation $x=\cos x$ is well-known because some facts. For example, with an old calculator, you can find approximations of the solution by typing any number and pressing the $\cos$ button repeatedly.

With Bolzano-Rolle combo, it is not difficult to show that this solution exists and is unique.

My question: Is there some work about the rationality or trescendality of this solution? Of course, $x$ is in radians.

You should think of it as intersection of two graphs: $ y=cos(x) \wedge y=x $. This allows determining if point (x,y) is inside the solution. — tp1, Oct 07 '22 at 01:59
@ajotatxe Cf. Is sin(x) necessarily irrational where x is rational?. — John Omielan, Oct 07 '22 at 02:00
$\cos x$ is the real part of $e^{ix}$. If $x$ is nonzero algebraic, then so is $ix$, so $e^{ix}$ is transcendental, so $\cos x$ is transcendental, so $\cos x\ne x$. See https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem — Gerry Myerson, Oct 07 '22 at 02:16

score 4 · Accepted Answer · answered Oct 07 '22 at 02:12

If the solution is algebraic, then both $x$ and $\cos x$ are algebraic. This implies $\sin x=\pm\sqrt{1-\cos^2x}$ is also algebraic. So $e^{ix}=\cos x+i\sin x$ is also algebraic, which is not possible for any non-zero algebraic $x$ by Lindemann-Weierstrass Theorem.

Hence the solution is transcendental.

About the algebraic character of the solution of $\cos x=x$

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