Let `S = { theta in [-2pi,2pi]:2 cos^(2) theta + 3 sin theta = 0}`, then the sum of the elements of S is .

Explanation

To solve the equation \( S = \{ \theta \in [-2\pi, 2\pi] : 2 \cos^2 \theta + 3 \sin \theta = 0 \} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2 \cos^2 \theta + 3 \sin \theta = 0 \]
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we can rewrite the equation as:
\[ 2(1 - \sin^2 \theta) + 3 \sin \theta = 0 \]
This simplifies to:
\[ 2 - 2 \sin^2 \theta + 3 \sin \theta = 0 \]
Rearranging gives:
\[ -2 \sin^2 \theta + 3 \sin \theta + 2 = 0 \]

### Step 2: Rearranging the equation
Multiplying through by -1, we get:
\[ 2 \sin^2 \theta - 3 \sin \theta - 2 = 0 \]

### Step 3: Factor the quadratic equation
Now we will factor the quadratic:
\[ (2 \sin \theta + 1)(\sin \theta - 2) = 0 \]
This gives us two equations to solve:
1. \( 2 \sin \theta + 1 = 0 \)
2. \( \sin \theta - 2 = 0 \)

### Step 4: Solve for \( \sin \theta \)
From the first equation:
\[ 2 \sin \theta + 1 = 0 \implies \sin \theta = -\frac{1}{2} \]
From the second equation:
\[ \sin \theta = 2 \]
Since the sine function cannot exceed 1, we discard this solution.

### Step 5: Find angles for \( \sin \theta = -\frac{1}{2} \)
The angles for which \( \sin \theta = -\frac{1}{2} \) in the interval \([-2\pi, 2\pi]\) are:
\[ \theta = -\frac{\pi}{6}, \quad \theta = -\frac{5\pi}{6}, \quad \theta = \frac{7\pi}{6}, \quad \theta = \frac{11\pi}{6} \]

### Step 6: Sum the elements of \( S \)
Now we will sum these angles:
\[ -\frac{\pi}{6} + -\frac{5\pi}{6} + \frac{7\pi}{6} + \frac{11\pi}{6} \]

Calculating the sum:
\[ -\frac{\pi}{6} - \frac{5\pi}{6} = -\frac{6\pi}{6} = -\pi \]
\[ \frac{7\pi}{6} + \frac{11\pi}{6} = \frac{18\pi}{6} = 3\pi \]
So the total sum is:
\[ -\pi + 3\pi = 2\pi \]

### Final Answer
Thus, the sum of the elements of \( S \) is:
\[ \boxed{2\pi} \]