Find six angles for which `sin theta=-(sqrt(3))/(2)`.

Answer

To find six angles for which \( \sin \theta = -\frac{\sqrt{3}}{2} \), we can follow these steps:

### Step 1: Identify the reference angle
The sine function is negative in the third and fourth quadrants. The reference angle for \( \sin \theta = \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{3} \).

### Step 2: Determine the angles in the third and fourth quadrants
1. In the third quadrant, the angle can be calculated as:
\[ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \]
2. In the fourth quadrant, the angle can be calculated as:
\[ \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \]

### Step 3: General solution for sine function
The general solution for \( \sin \theta = \sin \alpha \) is given by:
\[ \theta = n\pi + (-1)^n \alpha \]
where \( n \) is any integer and \( \alpha \) is the reference angle.

### Step 4: Apply the general solution
Here, \( \alpha = -\frac{\pi}{3} \). Therefore, we can write:
\[ \theta = n\pi + (-1)^n \left(-\frac{\pi}{3}\right) \]

### Step 5: Calculate specific angles for different values of \( n \)
1. For \( n = 0 \):
\[ \theta = 0 + (-1)^0 \left(-\frac{\pi}{3}\right) = -\frac{\pi}{3} \]
2. For \( n = 1 \):
\[ \theta = \pi + (-1)^1 \left(-\frac{\pi}{3}\right) = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \]
3. For \( n = 2 \):
\[ \theta = 2\pi + (-1)^2 \left(-\frac{\pi}{3}\right) = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \]
4. For \( n = 3 \):
\[ \theta = 3\pi + (-1)^3 \left(-\frac{\pi}{3}\right) = 3\pi + \frac{\pi}{3} = \frac{10\pi}{3} \]
5. For \( n = -1 \):
\[ \theta = -\pi + (-1)^{-1} \left(-\frac{\pi}{3}\right) = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3} \]
6. For \( n = -2 \):
\[ \theta = -2\pi + (-1)^{-2} \left(-\frac{\pi}{3}\right) = -2\pi - \frac{\pi}{3} = -\frac{7\pi}{3} \]

### Final angles
The six angles for which \( \sin \theta = -\frac{\sqrt{3}}{2} \) are:
1. \( -\frac{\pi}{3} \)
2. \( \frac{4\pi}{3} \)
3. \( \frac{5\pi}{3} \)
4. \( \frac{10\pi}{3} \)
5. \( -\frac{2\pi}{3} \)
6. \( -\frac{7\pi}{3} \)