The problem asks us to sketch the graph of the polar equation $r = 2\sin(2\theta)$.

1. Problem Description

The problem asks us to sketch the graph of the polar equation $r = 2\sin(2\theta)$.

2. Solution Steps

We need to analyze the polar equation $r = 2\sin(2\theta)$ to determine the shape of the graph.

First, let's consider some key values of $\theta$ and find the corresponding values of $r$:

- $\theta = 0$: $r = 2\sin(0) = 0$

- $\theta = \frac{\pi}{4}$: $r = 2\sin(\frac{\pi}{2}) = 2$

- $\theta = \frac{\pi}{2}$: $r = 2\sin(\pi) = 0$

- $\theta = \frac{3\pi}{4}$: $r = 2\sin(\frac{3\pi}{2}) = -2$

- $\theta = \pi$: $r = 2\sin(2\pi) = 0$

- $\theta = \frac{5\pi}{4}$: $r = 2\sin(\frac{5\pi}{2}) = 2$

- $\theta = \frac{3\pi}{2}$: $r = 2\sin(3\pi) = 0$

- $\theta = \frac{7\pi}{4}$: $r = 2\sin(\frac{7\pi}{2}) = -2$

- $\theta = 2\pi$: $r = 2\sin(4\pi) = 0$

The graph of $r = 2\sin(2\theta)$ is a four-leaf rose. Each leaf is symmetric about its axis.

- The first leaf is in the first quadrant, extending from $\theta = 0$ to $\theta = \frac{\pi}{2}$, with a maximum at $\theta = \frac{\pi}{4}$.

- The second leaf is in the second quadrant, extending from $\theta = \frac{\pi}{2}$ to $\theta = \pi$.

- The third leaf is in the third quadrant, extending from $\theta = \pi$ to $\theta = \frac{3\pi}{2}$.

- The fourth leaf is in the fourth quadrant, extending from $\theta = \frac{3\pi}{2}$ to $\theta = 2\pi$.

Each leaf has a maximum distance of 2 from the origin.

3. Final Answer

The graph of $r = 2\sin(2\theta)$ is a four-leaf rose, where each leaf has a maximum radius of 2.