To graph the polar equation \( r = 2 \sin(3\theta) \), we start by identifying the key components of the equation, which is in the form \( r = a \sin(n\theta) \). Here, \( a = 2 \) and \( n = 3 \). The value of \( n \) indicates the number of petals in the rose graph. Since \( n \) is odd, we will have exactly 3 petals.

Next, we plot the first petal. The angle for the first petal can be calculated using the formula \( \theta = \frac{\pi}{2n} \). Substituting \( n = 3 \), we find:

\( \theta = \frac{\pi}{2 \times 3} = \frac{\pi}{6} \).

At this angle, the radius \( r \) is equal to \( a = 2 \), so the first petal is located at the point \( \left(2, \frac{\pi}{6}\right) \).

To find the positions of the other petals, we need to determine the spacing between them. The petals are spaced apart by \( \frac{2\pi}{n} \). For our case:

\( \text{Spacing} = \frac{2\pi}{3} \).

Starting from the angle of the first petal \( \frac{\pi}{6} \), we add the spacing to find the next petal:

\( \frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{6} + \frac{4\pi}{6} = \frac{5\pi}{6} \).

At this angle, the radius remains \( r = 2 \), so the second petal is at \( \left(2, \frac{5\pi}{6}\right) \).

For the third petal, we add another \( \frac{2\pi}{3} \) to the angle of the second petal:

\( \frac{5\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6} + \frac{4\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \).

Thus, the third petal is located at \( \left(2, \frac{3\pi}{2}\right) \).

Now that we have the locations of all three petals, we can connect them with smooth curves. Remember that the petals will extend from the origin, creating a rose-like shape. The final graph of \( r = 2 \sin(3\theta) \) will exhibit three equally spaced petals, each extending out to a radius of 2.

This process illustrates how to graph polar equations of the rose type, emphasizing the relationship between the parameters \( a \) and \( n \) and the resulting geometric shape.