So, the equation is in the title. Is there a way to solve it in radicals or/and elementary functions?
It's an even function $f(x) = x \sin x$ with minimums and maximums not lying on lines $y=x$ and $y=-x$. I've tried to manipulate with its complex representation, but found nothing interesting about it.
On
Assuming that you may be concerned by the first zero of function $$f(x)=x \sin(x)-a$$ assuming $0 \leq a \leq \frac \pi 2$, what you could do is to use the usual series expansion that is to say $$f(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1)!}x^{2(n+1)}-a$$ and use series reversion to get something like $$x=\sqrt a \left(1+\frac{a}{12}+\frac{29 a^2}{1440}+\frac{263 a^3}{40320}+\frac{23479 a^4}{9676800}+O\left(a^5\right) \right)$$ Trying for a few values, some results $$\left( \begin{array}{ccc} a & \text{estimate} & \text{solution} \\ 0.1 & 0.318929 & 0.318929 \\ 0.2 & 0.455052 & 0.455053 \\ 0.3 & 0.562516 & 0.562517 \\ 0.4 & 0.655879 & 0.655886 \\ 0.5 & 0.740813 & 0.740841 \\ 0.6 & 0.820277 & 0.820357 \\ 0.7 & 0.896081 & 0.896278 \\ 0.8 & 0.969460 & 0.969899 \\ 0.9 & 1.041330 & 1.042232 \\ 1.0 & 1.112421 & 1.114157 \\ 1.1 & 1.183338 & 1.186529 \\ 1.2 & 1.254616 & 1.260274 \\ 1.3 & 1.326741 & 1.336512 \\ 1.4 & 1.400169 & 1.416772 \\ 1.5 & 1.475340 & 1.503412 \end{array} \right)$$ To polish the root, start Newton method with the estimate.
For example, for $a=1.5$, Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.4753402 \\ 1 & 1.5029586 \\ 2 & 1.5034118 \\ 3 & 1.5034119 \end{array} \right)$$
There are obvious "nice" solutions, specifically:
$a=0$ where $x=n\pi, n\in \mathbb{Z}$
and $a = (4n+1)\frac{\pi}{2},n\in \mathbb{Z}$ where $x= \pm a$
Apart from these there is probably no exact solution that can be expressed in terms of the standard elementary functions. This is as @EthanBolker has expressed in the comments.
I am not knowledgeable about the theory required to prove that, though.