Let $X=\mathbb{D}^2/\sim$, where $(\cos(\theta),\sin(\theta))\sim(\cos(\theta+\frac{2\pi}{3}),\sin(\theta+\frac{2\pi}{3}))$, $\theta\in \mathbb{R}$ Let $X=\mathbb{D}^2/\sim$, where $(\cos(\theta),\sin(\theta))\sim(\cos(\theta+\frac{2\pi}{3}),\sin(\theta+\frac{2\pi}{3}))$ for all $\theta\in \mathbb{R}$. How is this topological space graphically? What does it look like? Thank you.
 A: Notice that your equivalence relation is only defined for points on the boundary circle, so only those points are being identified. On the boundary $\partial D^2$, the equivalence class of a point $x = (cos \theta, sin \theta)$ is 
$$ [x] = \{\ (cos \theta, sin \theta),\  (cos (\theta + \frac{2\pi}{3}), sin (\theta + \frac{2\pi}{3}) ),\ (cos (\theta + \frac{4\pi}{3}), sin (\theta + \frac{4\pi}{3}) )\ \}$$
(This can be expressed more compactly by treating $x\in \partial D^2$ as a complex number: if we let $\xi$ be a primitive 3-rd root of unity then $[x] = \{ x, \xi x, \xi^2  x \}$. ) So on the boundary the quotient map takes the form $S^1 \to S$ where 
$$ S = \{ (cos \theta, sin \theta)\ |\ 0\leq \theta \leq \frac{2\pi}{3} \}/\left((1,0) \sim (cos\frac{2\pi}{3}, sin\frac{2\pi}{3})\right) $$
Then $S$ is homeomorphic to $S^1$ and the quotient map on the boundary has degree $3$.
In fact our quotient space $X = D^2/\sim$ is homeomorphic to the space obtained by attaching a disk $D^2$ to a circle $S^1$ via a map of the boundary $\partial D^2 \to S^1$ of degree 3 (exercise). This is called the "3-fold dunce cap" (at least that's what Munkres calls it in his algebraic topology book) and the construction generalizes to arbitrary $n\geq 1$. If you're familiar with polygonal representations of surfaces, you can express the $n$-fold dunce cap with an $n$-gon whose edges are all identified with the same orientation.  These are not embeddable in $\mathbb{R}^3$ unless $n=1$ (and as @pre-kidney points out in the comments we get $\mathbb{RP}^2$ if $n=2$) so it's not necessarily easy to visualize. 
(Note: there is a Wikipedia article about a dunce hat which is actually something slightly different, and the colliding nomenclature is extremely unfortunate.)

Edit: There was a discussion in the comments about how to compute the homology of this space, so I decided to include it in my answer for posterity (even though it's not technically part of the question). I will use the notation $q\colon D^2 \to X$ for the quotient map and $[x]$ the equivalence class of a single point, and I will stick to complex coordinates because they are more compact.
We can compute the cellular homology directly, given the description as a $2$-dimensional complex constructed by attaching $D^2$ to $S^1$ by a degree $3$ map on the boundary. If cellular homology is not available we can use Mayer-Vietoris instead.
Let $U = X \setminus [0]$ and $V = q(int D^2)$. Then $U$ and $V$ are open, $U\cup V = X$, and $U\cap V$ is homeomorphic to an open annulus and hence homotopy equivalent to $S^1$. Moreover $V$ is contractible and $U$ deformation retracts onto $S = q(\partial D^2)$, which is homeomorphic to $S^1$ as above. Then the interesting segment of the Mayer-Vietoris sequence is
$$\dots H_2(X) \to H_1(U\cap V) \to H_1(U) \oplus H_1(V) \to H_1(X) \to H_0(U\cap V) \to \dots $$
All of the groups to the left of this segment vanish, and the map $H_1(X) \to H_0(U \cap V)$ is $0$, so the sequence reduces to 
$$ 0 \to H_2(X) \to \mathbb{Z} \to \mathbb{Z} \to H_1(X) \to 0$$
and all that remains is to compute the homomorphism $H_1(U\cap V) \to H_1(U)$ induced by inclusion. We take it as given that if $Y\cong S^1$ then a generator of $H_1(Y)$ can be represented by any loop $\gamma \colon [0,1] \to Y$ that goes one time around. It follows then that a generator of $H_1(U\cap V)$ is given by $\alpha \colon [0,1] \to U\cap V$ where $\alpha(t) = [\frac{1}{2} e^{2\pi i t}]$, and a generator of $H_1(S)$ (and hence $H_1(U)$) is given by $\gamma\colon [0,1] \to S\subset U$ where $\gamma(t) = [e^{\frac{2\pi}{3} i t}]$. But in $U$, $\alpha$ is homotopic to $\tilde{\alpha}(t) = [e^{2\pi i t}]$ and they represent the same homology class, and we can see that $\tilde{\alpha} \sim \gamma * \gamma * \gamma$ (where $*$ is concatenation of paths). Therefore the map $H_1(U\cap V) \to H_1(U)$ is multiplication by $3$ and we have 
$$ H_0(X) \cong \mathbb{Z}, H_1(X) \cong \mathbb{Z}/3\mathbb{Z}, \text{ and }H_n(X) =0 \text { otherwise.} $$
