It is the set $V(y-\sin(x))\subset k^2$ a variety? It is the set $V(y-\sin(x))\subset k^2$ a variety?

I know that $V(y-\sin(x))=\{...,-4\pi,-3\pi,-2\pi,-\pi,0,\pi,2\pi,3\pi,4\pi,...\}$. Could it be a set of points a variety? My intuition tells me what not. How can I do a formal test, what property should I use? Thank you very much.
Suppose $X = V(y-\sin(x))\subset \Bbb A^2$ were a variety. Then $X$ also admits a description as $V(f_1,\cdots,f_n)$ for some finite list of nonzero polynomials $f_i(x,y)$, each of which vanishes on $V(y-\sin(x))$.

By examining $f_i(x,0)$, each single-variable polynomial $f_i(x,0)$ should vanish on $V(y-\sin(x))\cap V(y)$. On the other hand, as you (intended to) identify in your post, $V(y-\sin(x))\cap V(y)$ is infinite, which is a problem (Why? Try to identify for yourself before mousing over the following spoiler).


   Any nonzero polynomial in one variable may have at most finitely many solutions. As $f_i(x,0)$ has infinitely many solutions, it must be the zero polynomial, which implies that $y|f_i$ for each $f_i$, or that $V(y-\sin(x))$ contains the $x$-axis, which it does not by your identification of $V(y-\sin(x))\cap V(y)$.
$V=V(y-\sin(x))=\{(x,y); y=\sin(x) \}=\{(x,\sin(x)); x\in \Bbb R \}$  is not variety because no polynomial in $\Bbb R[x,y]$ has zeroes the set $V$.