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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.