Solve the equation. Give a general formula for all the solutions. List six solutions. \[ \begin{array}{l}\sin \theta=-\frac{\sqrt{3}}{2} \\ \text { Identify the general formula for all the solutions to } \sin \theta=-\frac{\sqrt{3}}{2} \\ \theta=\square, \mathrm{k} \text { is an integer }\end{array} \]

To solve the equation \(\sin \theta = -\frac{\sqrt{3}}{2}\), we need to find the angles where the sine function achieves that value. The sine function is negative in the third and fourth quadrants. The reference angle for \(\sin \theta = \frac{\sqrt{3}}{2}\) is \(\frac{\pi}{3}\) (or 60 degrees). Thus, in the third quadrant, the angle is \(\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\), and in the fourth quadrant, the angle is \(\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\). The general formulas for all solutions are: \[ \theta = \frac{4\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{3} + 2k\pi, \quad k \in \mathbb{Z} \] Now listing six specific solutions by varying \(k\): 1. For \(k = 0\): \(\theta = \frac{4\pi}{3}, \frac{5\pi}{3}\) 2. For \(k = 1\): \(\theta = \frac{4\pi}{3} + 2\pi = \frac{10\pi}{3}, \frac{5\pi}{3} + 2\pi = \frac{11\pi}{3}\) 3. For \(k = -1\): \(\theta = \frac{4\pi}{3} - 2\pi = -\frac{2\pi}{3}, \frac{5\pi}{3} - 2\pi = -\frac{\pi}{3}\) So the six solutions are: 1. \(\frac{4\pi}{3}\) 2. \(\frac{5\pi}{3}\) 3. \(\frac{10\pi}{3}\) 4. \(\frac{11\pi}{3}\) 5. \(-\frac{2\pi}{3}\) 6. \(-\frac{\pi}{3}\)