Finding the equivalence classes of a trigonometric relation

Start with the definition you have stated: $$[x] = \{ y \in \mathbb{R} \mid \sin(x) = \sin(y) \}\ .$$ I'm going to slightly change the notation: $$[a] = \{ x \in \mathbb{R} \mid \sin x = \sin a \}\ .$$ The reason I have done this is to emphasize what you have to do: given a fixed number $a$, find all $x$ which satisfy the equation. You should be able to do this by basic trigonometric methods, and the diagram and graph in your question ought to help. To give one example, $$\Bigl[\frac{\pi}{6}\Bigr] =\Bigl\{x\in\mathbb{R}\mid\sin x=\sin\Bigl(\frac{\pi}{6}\Bigr)\Bigr\} =\Bigl\{\frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi\mid k\in\mathbb{Z}\Bigr\}\ .$$

Hint: By basic trigonometry, $\sin(x)=\sin(x+2\pi)$.

Let $P(\alpha)=(x,y)$ be the point on the unit circle that corresponds to the angle $\alpha$. Then $\sin \alpha = y$. So if $\sin \alpha = \sin \beta$, then

$$\text{$\alpha = \beta + 2m\pi$ or $\alpha = \pi - \beta + 2n\pi$ for some $m,n \in \mathbb Z$}$$.

So $$[\theta] = (\theta + 2\pi m) \cup (\pi - \theta +2\pi n)$$ for some $m,n \in \mathbb{Z}.$