Draw and study the discontinuous of $g(x)=\lfloor \sin(x) \rfloor$
$$g(x)=\begin{cases} -1 & x\in (-\pi,0) \\ 0 & x \in [0,\pi]\setminus \lbrace \frac{\pi}{2} \rbrace \\ 1 & x=\frac{\pi}{2} \end{cases}.$$
I draw it like this:
We have that
$\lim_{x \to 0^-} g(x)=-1$ and $\lim _{x \to 0^+} g(x)=0$.
$\lim _{x \to {-\pi}^-} g(x)=0$ and $\lim _{x \to {-\pi}^+} g(x)=-1$
$\lim _{x \to {+\pi}^-} g(x)=0$ and $\lim _{x \to {+\pi}^+} g(x)=-1$
$\lim _{x \to {\frac{\pi}{2}}} g(x)=0$ but $f(\frac{\pi}{2})=1$
$\lim _{x \to {\frac{-3\pi}{2}}} g(x)=0$ but $f(\frac{-3\pi}{2})=1$
So, I think
The limit doesn’t exist if $\sin(x)=1$ or $\sin(x)=0$, this means that if $x=n \pi$ where $n\in \Bbb{Z}$, or if $x=2k \pi + \frac{\pi}{2}$ where $k \in \Bbb{Z}$
Therefore, the discontinuity points are $\lbrace k\pi, 2k \pi + \frac{\pi}{2} \mid k \in \Bbb{Z}\rbrace$.
Is that true, please?
Yes it is true that those are the discontinuities, you can use that $\sin(x)$ is a periodic function with period $2\pi$ to formalize your intuition.