Is it possible to find an exact solution for $x$ in $\frac{x^2+9}{6 x}=\frac{\sin (\pi -1) x}{\cos \pi x}$? I have this equation for $x>0$
$$\frac{x^2+9}{6 x}=\frac{\sin (\pi -1) x}{\cos \pi  x}$$
and I want to know if there is an xeact solution of $x$ in this equation.
 A: Here, I have a sort of "moral" argument for why there should be no poles. To begin, if there is a closed for for solutions to
$$\frac{(x^2+9)\cos(\pi x)}{x^2}=6\sin(\pi-1)$$
(which is equivalent to your problem) then we should expect that this neat formula will also give solutions to the equations
$$\frac{(x^2+9)\cos(\pi x)}{x^2}=c$$
for any constant $c$ since the constant $6\sin(\pi-1)$ is not nice at all and I cannot conceive of a formula that would solve this equation for only the value $c=6\sin(\pi-1)$. This means that we are looking for a "nice" function $f(c)$ such that
\begin{equation}
\frac{(f^2(c)+9)\cos(\pi f(c))}{f^2(c)}=c\tag{1}
\end{equation}
If $f(c)$ is a nice function, then surely a derivative of this function must exist (otherwise, I would not qualify $f(c)$ as a nice function). Thus,
$$\left(\frac{(f^2(c)+9)\cos(\pi f(c))}{f^2(c)}\right)^{'}=1$$
taking the derivative, we can then substitute in from (1) to write $\cos(\pi f(c))$
and $\sin(\pi f(c))$ in terms of a polynomial in $f(c),c$. Overall, this yields that
$$-18cf\left(c\right)^{-3}f'\left(c\right)=1+\frac{9}{f\left(c\right)^{2}}+\pi f'\left(c\right)\left(1+\frac{9}{f\left(c\right)^{2}}\right)^{2}\sqrt{1-\frac{c^{2}f\left(c\right)^{4}}{\left(f\left(c\right)^{2}+9\right)^{2}}}$$
Which, after expanding, is a polynomial in $f$ and $f'$. Apart from that, this differential equation has no special symmetry. It has been shown that general polynomials in $f$,$f'$, and $c$ have no closed form solution and so since we have no reason to expect that something special is going on here I would say that it is highly unlikely that your equation has a closed form. Not impossible, but definitely improbable.
