Claim: $(x)\neq \langle x \rangle$ in $\mathbb{R}[x].$
Notation:
$(x)$ is the principal ideal generated by x.
$\langle x \rangle$ is the same as in group theory, namely cyclic group generated by x.
My idea: Since the set of all polynomials with constant term $0$ is equal to $(x)$ and an ideal of $\mathbb{R}[x]$ (I have proved this) and then showing that $\langle x \rangle$ is not an ideal, the above claim is true.
However, I am confused on what $\langle x \rangle$ is in $\mathbb{R}[x]$. Is it just the set $\{x, 2x, 3x,..., nx\: \lvert n\in \mathbb{N}\}$? But isn't it generating the polynomial ring? What is it?
As you mentioned, $\langle x \rangle$ is the set $\{x, 2x, 3x,...,nx,... \: \lvert n\in \mathbb{N}\}$. But $(x) = \{rx \lvert r\in \mathbb{R}[x]\}$. Then you can see that $x^2 \in (x)$,but $x^2\notin \langle x \rangle$ as mentioned in the comment. In general, $\langle x \rangle \subseteq (x) $.