Suppose `cos x = 0` and `cos(x+z)=1/2`. Then, the possible value (s) of `z` is (are).

Options

A. (a) `pi/6`
B. (b) `(5pi)/(6)`
C. (c) `(7pi)/(6)`
D. (d) `(11pi)/(6)`

Answer

To solve the problem, we start with the given conditions:

1. \( \cos x = 0 \)
2. \( \cos(x + z) = \frac{1}{2} \)

### Step 1: Analyze \( \cos x = 0 \)

The cosine function is zero at specific angles. The general solutions for \( x \) where \( \cos x = 0 \) are:
\[ x = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 2: Substitute \( x \) into the second equation

Given \( \cos(x + z) = \frac{1}{2} \), we can use the cosine addition formula:
\[ \cos(x + z) = \cos x \cos z - \sin x \sin z \]
Substituting \( \cos x = 0 \):
\[ 0 \cdot \cos z - \sin x \sin z = \frac{1}{2} \]
This simplifies to:
\[ -\sin x \sin z = \frac{1}{2} \]
Thus,
\[ \sin x \sin z = -\frac{1}{2} \]

### Step 3: Determine \( \sin x \)

Since \( \cos x = 0 \), we can find \( \sin x \):
\[ \sin^2 x + \cos^2 x = 1 \implies \sin^2 x + 0 = 1 \implies \sin x = \pm 1 \]
Thus, \( \sin x = 1 \) or \( \sin x = -1 \).

### Step 4: Case 1: \( \sin x = 1 \)

Substituting \( \sin x = 1 \) into the equation:
\[ 1 \cdot \sin z = -\frac{1}{2} \implies \sin z = -\frac{1}{2} \]
The solutions for \( \sin z = -\frac{1}{2} \) are:
\[ z = \frac{7\pi}{6} + 2k\pi \quad \text{and} \quad z = \frac{11\pi}{6} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]

### Step 5: Case 2: \( \sin x = -1 \)

Substituting \( \sin x = -1 \) into the equation:
\[ -1 \cdot \sin z = -\frac{1}{2} \implies \sin z = \frac{1}{2} \]
The solutions for \( \sin z = \frac{1}{2} \) are:
\[ z = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad z = \frac{5\pi}{6} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]

### Step 6: Compile all possible values of \( z \)

From both cases, we have the following possible values for \( z \):
1. \( z = \frac{\pi}{6} \)
2. \( z = \frac{5\pi}{6} \)
3. \( z = \frac{7\pi}{6} \)
4. \( z = \frac{11\pi}{6} \)

### Final Answer

The possible values of \( z \) are:
\[ z = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \]

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