It's easy to show (using the Lindemann-Weierstrass theorem) that, for $x\ne 0$, at least one of $x$ and $\sin(x)$ must be transcendental.
But what about $x\sin(x)$?
After all, the product of two transcendental numbers could be algebraic. Hence: Can the product $x\sin(x)$ be non-zero and algebraic? Might it even be rational? (The requirement that $x\sin(x)\ne 0$ is meant to rule out the trivial case of $x=k\pi$.)
Best Answer
By request, this is my comment, promoted to an answer:
$\lim\sup_{x\to\infty} x \sin(x)=\infty$ and $\lim\inf_{x\to\infty} x \sin(x)=-\infty$, so by continuity $x \sin(x)$ takes any real value, in particular all rational and all algebraic ones.