Ian Morris has it essentially correct in his comment. If you solve the equation $x = (\sin \pi y/2)^2$ for $y$, then the orbit of $y \in \mathbb{R}/\mathbb{Z}$ under $y \mapsto 2y$ is dense if and only if $y$ is 2-normal. In other words, if every finite binary string appears in the binary expansion of $y$. If you look at the Wikipedia page for normal numbers, it's clear that there won't be a simpler description than that of any proven example. You can certainly conjecture that any reasonably simple choice of $x$ that makes $y$ irrational also makes $y$ 2-normal, but such numbers have generally not been proven to be normal.

Ian Morris has it essentially correct in his comment. If you solve the equation $x = (\sin \pi y/2)^2$ for $y$, then the orbit of $y \in \mathbb{R}/\mathbb{Z}$ under $y \mapsto 2y$ is dense if and only if every finite binary string appears in the binary expansion of $y$. Now, this is weaker than being normal in base 2, because that requires that every binary string appears equally often, not just that it appears. Let's call such a number "topologically 2-normal" (or 2-dense could be another name), because the 2-normality condition is equivalent to saying that the orbit of $y$ is not just dense, but ergodic. My impression is that not much more is known about topologically normal numbers than about normal numbers. For instance, you can conjecture that any irrational algebraic number is topologically normal in base 2 (or in any other base), but it doesn't look like it is known. In any case, topological normality is the heart of the question.