Is the root of $x=\cos(x)$ a transcendental number?

This question struck me when thinking about the fixed point of $x=\cos(x)$ being "obviously" not an algebraic number (unlike something like $\sqrt{2}$, see this question).

If so, how would one prove this?

I don't see how something like a simple application of the Lindemann-Weierstrass Theorem would work in this case.

I understand that the mere fact that it is a root of a so-called transcendental equation doesn't amount to very much, since $x=\tan(x)$ has at least one root (0) which is algebraic.

The Lindemann-Weierstrass Theorem states that given $\alpha_1,\ldots, \alpha_n$ algebraic numbers that are linearly independent over $\mathbb{Q}$, then $e^{\alpha_1},\ldots, e^{\alpha_n}$ are algebraically independent over $\mathbb{Q}$ (i.e. for any rational function $P(z_1,\ldots, z_n)$ with algebraic coefficients it follows that $P(e^{\alpha_1},\ldots, e^{\alpha_n})\neq 0$). As a result, with $n=1$ we find that if $\alpha$ is algebraic that is non-zero, then $e^\alpha$ is transcendental.

Now, suppose that $a$ is algebraic; then $ia$ is also algebraic because $i$ is algebraic and the set of all algebraic numbers is a field. Next, consider the rational function $P(z)=\frac{z^2-(2\cos a) z + 1}{2z}$ and assume for the sake of a contradiction that $\cos a$ is algebraic. $P(z)$ is not identically zero and has algebraic coefficients, but $P(e^{ia})=0$ by the definition $\cos a=\frac{e^{ia}+e^{-ia}}{2}$, contradicting the assumption that $a$ is algebraic by Lindemann-Weierstrass.

Thus, we find that if $a$ is algebraic, then $\cos a$ is transcendental. As a result, if $a$ is a complex number such that $a=\cos a$, then if $a$ is algebraic, $\cos a=a$ is transcendental, giving a contradiction. Since $a\neq 0$ because $\cos 0 = 1$, we find that $a$ is transcendental.