Algebraic Solution to $\cos(\pi x) + x^2 = 0$ Today I was fiddling about with a TI-89 calculator, attempting as usual to confuse it. I figured that making it solve an equation with a periodic function would be fun, so I tried the following:
$$\cos(\pi x) + x^2 = 0$$
While this didn't stump it, I noted that the solution it gave was just a decimal. Since I ran this through solve(), which usually will give you a nice solution like $\frac{\sqrt{2}}{2}$ if it can, I found this rather interesting. I figured what it eventually did was give up on solving it in its usual manner and move to a numeric method: i.e., it simply made intelligent guesses until it found good solutions.
I am usually inclined to think most random decimals have a closed form expression behind them. (In fact, do all such decimals have a closed form, even if we don't know it? I may have to look into that.) As such, I decided to put this into Wolfram|Alpha and see if it had any better results. But no dice; it gave me back, similar to the TI-89, that $x = \pm 1$ and $x \approx \pm 0.629847$ were solutions. 
The latter decimal, $x \approx 0.629847$, is the one I am concerned with. As far as I am concerned, $x = \pm 1$ are sort of "trivial" solutions; just thinking the problem over leads you to them naturally. 
Is there a way to solve this algebraically? I can sort of narrow it down. I know that as $x \to \infty$, the $\cos(\pi x)$ term is essentially trivial compared to $x^2$. Given that $x^2>0$ for any $x \ne 0$, and given that $\cos(\pi x)$ has a range of $[-1, 1]$, it seems to me that whenever $x^2>1$, $\cos(\pi x)$ cannot pull down $x^2$ enough for it become zero.
So, it seems natural to me to think, then, that all solutions must lie where $x^2 \le 1$, viz., within the interval $[-1, 1]$. This narrows down the field significantly, but it still does not really help me with an algebraic solution. (However, I figure if I were to come across this in a real-world scenario, this would be a useful line of attack for a guess-and-test sort of deal.)
Another line of attack I attempted was to take the reverse approach: use the numerically attained solution to find a closed-form solution. I thought it may be an interesting number I simply had not learned about, so I tried to look up the decimal sequence in the OEIS, but to no luck: no such sequence was available. 
I've sort of rambled, so here are my questions:


*

*Is there an algebraic solution to the above equation?

*Even if there is not, is there any way to figure out the closed form expression behind the decimal $x = 0.629847$? I don't even care if the expression has $\cos$ or $\sin$ in it. 


I will be honest: I really don't even know where to start. 
 A: Your $x \approx 0.629847$ is transcendental. First, $x \neq 0, \; 1/3, \; 2/3, \; 1 \;$ so by Corollary 3.12 in Niven, Irrational Numbers, page 41, $x$ is irrational. Now, one value of 
$$  (-1)^x  $$
is
$$  e^{i \pi x} = \cos \pi x + i \sin \pi x.   $$ As $x$ is not rational, Gelfond-Schneider, Theorem 10.1 on page 134, $  (-1)^x  $ is transcendental. 
The algebraic numbers in $\mathbb C$ are a field containing $i,$ the rationals, and closed under complex conjugation. It follows that a number is algebraic if and only if both its real and imaginary parts are algebraic. From $\cos^2 \pi x + \sin^2 \pi x = 1,$ it follows that either both parts are algebraic or both are transcendental. Therefore $\cos \pi x$ is transcendental. Since $x^2 = - \cos \pi x,$ we find that $x$ itself is transcendental.
Meanwhile, lots of nice numbers are transcendental. $\pi, \; e, \; \log 2 \; $ are transcendental but would be considered pleasant answers to a problem of this type. It is very, very hard to show that a number such as your $x$ does not have a nice closed form expression. I can't see how it could, of course.
A: The inverse symbolic calculator gives a few dozen numbers starting with .629847; maybe if you could get a couple more decimal places you'd narrow it down. Of course, Will is right that it's unlikely to have a simple expression in terms of familiar constants such as $\pi$, $\sqrt2$, etc., but it still might be a value of some special function someone has tabulated. 
A: Fourier Series 1:
The Fourier sine series of the inverse of $\cos(\pi x)+x^2-1$ converges on $0\le x\le\frac{19}{18},0\le y\le \frac23$:

Plugging in $x=1$ into the series expansion gives:
$$\cos(\pi x)+x^2=0\implies x=-\frac{36}{19}\sum_{n=1}^\infty\sin\left(\frac{18\pi n}{19}\right)\int_0^\frac23t(\pi\sin(\pi t)-2t)\sin\left(\frac{18\pi n}{19}(\cos(\pi t)+t^2-1)\right)dt$$
The inverse function is monotonic and therefore of bounded variation. A bounded variation function Fourier series converges everywhere. Therefore, this series solution  converges slowly to $0.6429\dots$ even though we input $x=1\lesssim \frac{19}{18}$
Fourier Series 2:
A slightly faster, but more complicated series of the inverse of $\cos\left(\pi\left(x+\frac12\right)\right)+\left(x+\frac12\right)^2-\frac14$ converges on $0\le x\le \frac 5{16}-\frac1{\sqrt2},0\le y\le \frac14$. The following graph shows an approximation of $f^{-1}(f(x))=x$:

Now $x=\frac14$
$$\cos(\pi x)+x^2=0\implies x=\frac{32}{8\sqrt2-5}\sum_{n=1}^\infty\sin\left(\frac{4\pi n}{8\sqrt2-5}\right)\int_0^\frac14t(\pi\cos(\pi t)-2t-1)\sin\left(\frac{16\pi n(\sin(\pi t)-t^2-t)}{8\sqrt2-5}\right)dt$$
Shown here
Now to integrate maybe with Bessel functions or a series expansion for 2 explicit series solutions.
