What is the solution of $\cos(x)=x$? There is an unique solution with $x$ being approximately $0.739085$. But is there also a closed-form solution?
 A: As I know, there is no exact way to get the solution for $\cos(x)=x$. But, you can use Newton's Method to get an approximate answer:

Consider the function $f(x)=\cos x−x$
This gives us that $f'(x)=-\sin x-1$

Newton's Method states that $x_{\text{n+1}} = x_{\text{n}} - \dfrac{f(x_{\text{n}})}{f'(x_{\text{n}})}$
Then, just start at $x_{\text{0}}=1$ and repeat this method all over again, until you are satisfied. Don't forget that rounding numbers might lead to wrong answers!
A: Here is a closed form I just found. Using Kepler’s Equation, the median of a beta distribution, and Inverse Beta Regularized $\text I^{-1}_z(a,b)$. Here is the closed form:
$$\boxed{\text{Dottie Number}=\text D=\sin^{-1}\left(1-2\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)\right)}$$
from The Incomplete Beta function $\text B_z(a,b)$:
$$\text B_{\sin^2(z)}\left(\frac 12,\frac32\right)=z+\cos(z)\sin(z)$$
taking the inverse, and a bit of algebra to get the form above. Also using the Half Covered Sine:
$$\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)=\text{hacoversin}(\text {D)}=\frac12(1-\sin(\text D))\implies \text D= {\text{hacoversin}}^{-1} \text I^{-1}_\frac12\left(\frac 12,\frac 32\right) $$
With an error of $10^{-179}$ in this numerical evaluation. This is a side post since there already is an accepted answer. Please correct me and give me feedback!

See explanation here

A: It is very easy to show that the equation $\cos x = x$ has a unique solution. For example take $f(x) = x - \cos x$ and notice that $f'(x) = 1+\sin x \ge 0$ (equality holding in isolated points) so $f(x)$ is strictly increasing and hence the equation can have at most one solution. Since $f(x)>0$ for $x\ge 1$ and $f(x)<0$ for $x\le 0$, and the function is continuous, by the intermediate values theorem there exists one and only one solution $\bar x \in [0,1]$.
For this particular equation there is also a very nice numeric approximation. In fact $\bar x = \lim x_n$ where $x_{n+1} = \cos (x_n)$ is any iteration of the function $\cos x$. You can easily find the numeric value for $\bar x$ simply putting any number in your pocket calculator and pressing repeatedly the $\cos$ button. In fact $\bar x$ is the fixed point of the $\cos$ function and, (at least in $[0,1]$) the $\cos$ function is a contraction hence every iterated sequence converges to the unique fixed point.
I can also convince you that $\bar x$ is an exact solution to the equation $\cos x = x$. I think that you agree that $\sqrt[3]2$ is an exact solution of the equation $x^3=2$, don't you? Now notice what's going on here... one notices that the function $x^3$ is strictly increasing hence invertible. You give a name to the inverse function and call it: cubic root. Then you find an algorithm to compute the cubic root on your calculator. Isn't this the same thing we did with the function $f(x) = x-\cos x$? 
By definition the number $q = \sqrt[3]2$ is the only real number such that $q^3=2$. Analogously, the number $\bar x$ is the only number such that $\bar x-\cos \bar x=0$.
A: 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 $f(x)=f(\cos x)\ \forall x \in \Bbb{R}$.
A: I'm truth, there the above answers are correct, stating that the number currently has no closed form solution.
An old approximation that I came up with to $\cos$ can offer a decent approximation to the Dottie number however.
$$\sqrt[3.911]{\frac{1-x^2}{1+x^2}}
\approx \cos{x}$$
Setting the above approximation $=x$ and solving can give a closed form solution that lies within 4 decimal points of the solution.
A: Mathworld calls this the Dottie Number. This page mentions a closed form
$$\sqrt{1-\left(2I_{\frac{1}{2}}^{-1}\left(\frac{1}{2},\frac{3}{2}\right)-1\right)^2}\,,$$
where $I_z^{-1}(a,b)$ is the inverse of the regularized beta function.
A: 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 $0 < x < \pi/2$
A: Your equation cannot be solved in terms of elementary functions, elementary functions and Lambert W or elementary functions and Generalized Lambert W of Mezö et al.. It can be solved in terms of "Leal-functions" and possibly by Generalized Lambert W of [Castle 2018].
1.) Elementary functions, elementary numbers
$$\cos(x)=x$$
$$\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix}=x$$
$$\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix}-x=0$$
$$\frac{1}{2}(e^{ix})^2-xe^{ix}+\frac{1}{2}=0$$
$x\to\frac{t}{i}$:
$$\frac{1}{2}(e^t)^2+ite^t+\frac{1}{2}=0$$
The function on the left-hand side of the latter equation is an algebraic function in dependence of both $t$ and $e^t$. Liouville proved that such kind of functions (over a complex domain without isolated points) don't have (partial) inverses that are elementary functions.
The equation is also an algebraic equation in dependence of both $t$
and $e^t$. Lin proved, assuming Schanuel's conjecture is true, that such kind of equations don't have solutions except $0$ that are elementary numbers.
How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?
2.) Lambert W, Generalized Lambert W
The latter equation also shows that the equation cannot be solved in terms of elementary functions and Lambert W or Generalized Lambert W of Mezö et. al. either. But possibly is it solvable in terms of Generalized Lambert W of [Castle 2018].
[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018
3.) "Leal-functions"
$$\cos(x)=x$$
$$\cos(x)-x=0$$
$x\to-t$:
$$t+\cos(t)=0$$
$$t=\text{Lcos}_2(0)$$
$$x=-\text{Lcos}_2(0)$$
[Vazquez-Leal et al. 2020] Vazquez-Leal, H.; Sandoval-Hernandez, M. A.; Filobello-Ninoa, U.: The novel family of transcendental Leal-functions with applications to science and engineering. Heliyon 6 (2020) (11) e05418
A: You could use cosine Taylor expansion and solve the polynomial 
$\sum_{i=0}^{n}(-1)^i \frac{x^{2i}}{(2i)!}-x=0$
For example; for n=1,2 you get the approximations $x=1,x=\sqrt{3}-1$.
A: I would say that it IS already in closed form... if we follow the definition given here :https://en.m.wikipedia.org/wiki/Closed-form_expression
since trigonometric functions are considered "well-known"... don't you agree?
A: My AskConstants "constant recognition" program at http://AskConstants.org proposed the explicit exact closed form
    RealInverseSphericalBesselY [0, -1, 1],

where $0$ is the order and $1$ is the branch number.
This was subsequently proved in my article at https://arxiv.org/abs/2207.00707
A: this can be really easy with the help of graphs. Draw the graphs of y= cosx and x=y. Intersection points on the graph will show the solutions.
