If `alpha and beta` are the solution of `sinx=-(1)/(2)` in `[0, 2pi]` and `alpha and gamma` are the solutions of `cos x=-(sqrt3)/(2)` in `[0, 2pi]`, then the value of `(alpha+beta)/(|beta-gamma|)` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the values of \( \alpha \), \( \beta \), and \( \gamma \) based on the given equations and then compute the expression \( \frac{\alpha + \beta}{|\beta - \gamma|} \).

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

The solutions for \( \sin x = -\frac{1}{2} \) in the interval \( [0, 2\pi] \) are found in the third and fourth quadrants.

1. The reference angle for \( \sin x = \frac{1}{2} \) is \( \frac{\pi}{6} \).
2. In the third quadrant:
\[ x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \]
3. In the fourth quadrant:
\[ x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \]

Thus, we have:
- \( \alpha = \frac{7\pi}{6} \)
- \( \beta = \frac{11\pi}{6} \)

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

The solutions for \( \cos x = -\frac{\sqrt{3}}{2} \) in the interval \( [0, 2\pi] \) are found in the second and third quadrants.

1. The reference angle for \( \cos x = \frac{\sqrt{3}}{2} \) is \( \frac{\pi}{6} \).
2. In the second quadrant:
\[ x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \]
3. In the third quadrant:
\[ x = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \]

Thus, we have:
- \( \alpha = \frac{7\pi}{6} \) (already found)
- \( \gamma = \frac{5\pi}{6} \)

### Step 3: Calculate \( \frac{\alpha + \beta}{|\beta - \gamma|} \)

Now we can substitute the values of \( \alpha \), \( \beta \), and \( \gamma \) into the expression.

1. Calculate \( \alpha + \beta \):
\[ \alpha + \beta = \frac{7\pi}{6} + \frac{11\pi}{6} = \frac{18\pi}{6} = 3\pi \]

2. Calculate \( |\beta - \gamma| \):
\[ \beta - \gamma = \frac{11\pi}{6} - \frac{5\pi}{6} = \frac{6\pi}{6} = \pi \]
Thus, \( |\beta - \gamma| = \pi \).

3. Now substitute into the expression:
\[ \frac{\alpha + \beta}{|\beta - \gamma|} = \frac{3\pi}{\pi} = 3 \]

### Final Answer
The value of \( \frac{\alpha + \beta}{|\beta - \gamma|} \) is \( 3 \).

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