I don't have any trouble showing that for any function $f: X \to Y$, the relation $x \sim y$ if and only if $f(x) = f(y)$ is an equivalence relation, all properties of which follow from reflexivity, symmetry, and transitivity of equality.
Now, I take $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sin(x)$, so $x \sim y$ if and only if $\sin(x) = \sin(y)$. This is certainly an equivalence relation, so I need only describe the equivalence classes and quotient set. I'm not sure I've done so accurately. Here is my (updated, after some helpful comments) attempt.
Let $x \in \mathbb{R}$. We have $x \sim y$ if and only if $\sin(x) = \sin(y)$, so the equivalence class of $x$ is the fibre of $x$ over the sine map (and the quotient set is the set of all such fibres). In particular, we know that $\sin(x) = \sin(x + 2\pi k)$ ($k \in \mathbb{Z}$) and $\sin(x) = \sin(\pi - x)$, so $\sin(x) = \sin(\pi - x) = \sin(\pi - x + 2\pi k)$. We then have: \begin{align*} x \sim y & \iff \sin(x) = \sin(y) \\ & \iff y = x + 2\pi k, k \in \mathbb{Z} \text{ or } y = \pi - x + 2\pi k, k \in \mathbb{Z} \\ & \iff y = x + 2k \pi, k \in \mathbb{Z} \text{ or } y = (2k + 1)\pi - x, k \in \mathbb{Z} \\ & \iff y = 2k \pi + (-1)^{2k} x, k \in \mathbb{Z} \text{ or } y = (2k+1) \pi + (-1)^{2k+1} x \\ & \iff y = n \pi + (-1)^n x, n \in \mathbb{Z}. \end{align*} So $$ [x] = \{y \in \mathbb{R} \mid \exists n \in \mathbb{Z}, \; y = n \pi + (-1)^n x\}. $$
How does this updated attempt look? The only "description" of the quotient set that I know of, other than the set of all equivalence classes $[x]$ above, is the set of fibres of the sine map.