Let $\pi : E \to X$ be a smooth rank $k$ vector bundle on a manifold $X$ (I don't think my question depends upon the stipulations on the bundle, but I've just chosen smooth in case I'm incorrect). By definition, for every $x \in X$, there exists an open set $U \subseteq X$, with $x \in U$, such that $\pi^{-1}(U)$ is diffeomorphic to $U\times\mathbb{R}^k$; we say that $U$ trivialises the bundle.
The prototypical example of a smooth vector bundle on a manifold $X$ is the tangent bundle $TX$. For this bundle (in fact any tensor bundle), any coordinate neighbourhood on $X$ trivialises the bundle. My question is whether this happens for every vector bundle.
Let $\pi : E \to X$ be a smooth rank $k$ vector bundle on a manifold $X$. Does every coordinate neighbourhood on $X$ trivialise the bundle?
The only way I think this can fail is if there is a bundle such that, for a given coordinate neighbourhood $U$, you must pass to a smaller open set $U' \subset U$ in order to trivialise the bundle.