limits - Does the Squeeze Theorem apply to $lim_{xtoinfty}sin(frac{pi x}{2-3x})$?

I was recently asked to find the limit of
$$\lim_{x\to\infty}\sin(\frac{\pi x}{2-3x}).$$
I eventually solved it by breaking the limit down to a composition of functions, but my first instinct was actually to invoke the Squeeze Theorem. My logic was that since

$$-1\leq \sin(x) \leq 1,$$
I can have $\lim_{x\to\infty}-1$ serve as the lower bound and $\lim_{x\to\infty}1$ serve as the upper bound to conclude that
$$\lim_{x\to\infty}-1=\lim_{x\to\infty}1=\lim_{x\to\infty}\sin(\frac{\pi x}{2-3x})=1.$$
Given that the second method I did yielded an answer of $-\frac{\sqrt{3}}{2}$, I'm not sure where I went wrong in my implementation of the Squeeze Theorem. My suspicion is that a) I was wrong in thinking $\frac{\pi x}{2-3x}$ would be akin to simply another $x$ if both were entered into $\sin(x)$, or b) I needed to break down the limit into compositions before invoking the Squeeze Theorem.