This is not quite the same as the old (closed) problem. For each $a$, the equation $x\sin(x)=a$ has solutions near $n\pi$ for every sufficiently large integer $n$. For example when $a=3$ as in this thread, we need only exclude $n=0,\pm 1$; for every other $n$, Newton's Method will converge quickly to a numerical solution of the OP's problem.
But this value $a=3$ is big enough to exclude another sort of solution that could have been offered in the old thread. Because as long as $0<a< 1.8197$ or so (the height of the first local max) it also makes sense to ask whether we can solve $x\sin(x)=a$ as a smooth function $x=f(a)$. The answer is yes, and although there's no handy "symbolic" expression for $f$, there is a power-series solution:
$$f(a) = b + (1/12) b^3 + (29/1440) b^5 + (263/40320) b^7 + \dots$$
where $b$ is either of the square roots of $a$. (The coefficients are obtained by inverting the power series of $x\sin(x)$ and then extracting a square root. I don't see any particular pattern to them; they're rational and once multiplied by an appropriate factorial they become nearly integral; they all seem to be positive and decreasing at a rate that suggests the appropriate radius of convergence.)