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 $x=k\pi$.)
On
Complementing ( & Complimenting ! ) the Pertinent Comment by "David Gao" , I am giving my matching observations , intuitions & thoughts here.
Consider the Curves $y=x$ , $y=\sin(x)$ , $y=x\sin(x)$ , generated courtesy of wolfram online tool & then modified a little :
Consider the Purple line $y=0.5$ , which intersects $y=x\sin(x)$ , hence we have rational value there. It must correspond to transcendental $x$ & transcendental $\sin(x)$ , though the Product is rational.
We can freely move the Purple line up or down to some other rational number or algebraic number like $1$ or $\sqrt{2}$ or $-\sqrt[3]{3}$. We will still have the Product of 2 transcendental numbers generating a Point on $x\sin(x)$ which is not a transcendental number.
ADDENDUM :
(1) We know , via the given LW Theorem , that we must have at least one of $x$ & $\sin(x)$ to be transcendental. More-over , because Product of transcendental & rational (algebraic) will be transcendental , Contradicting the rational (algebraic) value we selected , we know that we must have both transcendental $x$ & transcendental $\sin(x)$ here.
(2) Examples and arguments use the values between $+1$ & $-1$ because that is the range of $\sin$ , though that restriction is not necessary. We can easily see that at arbitrarily large $x$ values , we will have arbitrarily large $x\sin(x)$ , when-ever $|\sin(x)|$ goes to $1$.
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.