What is the general value of `theta` which satisfies both the equations `sin theta =- (1)/(2) and cos theta = - (sqrt3)/(2)` ?

Answer

To solve the problem of finding the general value of \( \theta \) that satisfies both equations \( \sin \theta = -\frac{1}{2} \) and \( \cos \theta = -\frac{\sqrt{3}}{2} \), we will follow these steps:

### Step 1: Solve \( \sin \theta = -\frac{1}{2} \)

We know that \( \sin \theta = -\frac{1}{2} \) corresponds to specific angles in the unit circle. The reference angle for \( \sin \theta = \frac{1}{2} \) is \( \frac{\pi}{6} \). Since we are looking for where the sine is negative, we find the angles in the third and fourth quadrants:

\[ \theta = 2n\pi + \left(-\frac{\pi}{6}\right) \quad \text{(fourth quadrant)} \]
\[ \theta = (2n + 1)\pi + \frac{\pi}{6} \quad \text{(third quadrant)} \]

Thus, we can express the solutions as:
\[ \theta = 2n\pi - \frac{\pi}{6} \quad \text{and} \quad \theta = (2n + 1)\pi + \frac{\pi}{6} \]

### Step 2: Solve \( \cos \theta = -\frac{\sqrt{3}}{2} \)

Next, we solve \( \cos \theta = -\frac{\sqrt{3}}{2} \). The reference angle for \( \cos \theta = \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{6} \). The angles where cosine is negative are in the second and third quadrants:

\[ \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \quad \text{(second quadrant)} \]
\[ \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad \text{(third quadrant)} \]

Thus, we can express the solutions as:
\[ \theta = 2n\pi + \frac{5\pi}{6} \quad \text{and} \quad \theta = 2n\pi + \frac{7\pi}{6} \]

### Step 3: Find common solutions

Now, we need to find \( \theta \) values that satisfy both equations. We have:

1. From \( \sin \theta = -\frac{1}{2} \):
- \( \theta = 2n\pi - \frac{\pi}{6} \)
- \( \theta = (2n + 1)\pi + \frac{\pi}{6} \)

2. From \( \cos \theta = -\frac{\sqrt{3}}{2} \):
- \( \theta = 2n\pi + \frac{5\pi}{6} \)
- \( \theta = 2n\pi + \frac{7\pi}{6} \)

To find common solutions, we can set \( \theta = 2n\pi + \frac{5\pi}{6} \) equal to \( \theta = (2m + 1)\pi + \frac{\pi}{6} \) for integers \( n \) and \( m \).

### Step 4: Check for integer values

1. Setting \( 2n\pi + \frac{5\pi}{6} = (2m + 1)\pi + \frac{\pi}{6} \):
- Rearranging gives \( 2n\pi - (2m + 1)\pi = -\frac{4\pi}{6} \)
- This simplifies to \( (2n - 2m - 1)\pi = -\frac{2\pi}{3} \)
- Thus, \( 2n - 2m - 1 = -\frac{2}{3} \) which does not yield integer solutions.

2. Setting \( 2n\pi + \frac{7\pi}{6} = (2m + 1)\pi + \frac{\pi}{6} \):
- Rearranging gives \( 2n\pi - (2m + 1)\pi = -\frac{6\pi}{6} \)
- This simplifies to \( (2n - 2m - 1)\pi = -1\pi \)
- Thus, \( 2n - 2m - 1 = -1 \) leading to \( 2n - 2m = 0 \) or \( n = m \).

### Final General Solution

The common solution can be expressed as:
\[ \theta = 2n\pi + \frac{7\pi}{6} \]

Thus, the general solution for \( \theta \) that satisfies both equations is:
\[ \theta = 2n\pi + \frac{7\pi}{6}, \quad n \in \mathbb{Z} \]