Number of values of x satisfying the equation `cos ( 3 arc cos ( x-1)) = 0 ` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation \( \cos(3 \arccos(x-1)) = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos(3 \arccos(x-1)) = 0 \]
This implies that:
\[ 3 \arccos(x-1) = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 2: Solve for \( \arccos(x-1) \)
Dividing both sides by 3 gives:
\[ \arccos(x-1) = \frac{\pi}{6} + \frac{n\pi}{3} \]

### Step 3: Apply the cosine function
Now, we apply the cosine function to both sides:
\[ x - 1 = \cos\left(\frac{\pi}{6} + \frac{n\pi}{3}\right) \]
Thus,
\[ x = 1 + \cos\left(\frac{\pi}{6} + \frac{n\pi}{3}\right) \]

### Step 4: Calculate values for different \( n \)
Now we will calculate \( \cos\left(\frac{\pi}{6} + \frac{n\pi}{3}\right) \) for different integer values of \( n \):

- **For \( n = 0 \)**:
\[ x = 1 + \cos\left(\frac{\pi}{6}\right) = 1 + \frac{\sqrt{3}}{2} = 1 + \frac{\sqrt{3}}{2} \]

- **For \( n = 1 \)**:
\[ x = 1 + \cos\left(\frac{\pi}{6} + \frac{\pi}{3}\right) = 1 + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \]

- **For \( n = 2 \)**:
\[ x = 1 + \cos\left(\frac{\pi}{6} + \frac{2\pi}{3}\right) = 1 + \cos\left(\frac{5\pi}{6}\right) = 1 - \frac{\sqrt{3}}{2} = 1 - \frac{\sqrt{3}}{2} \]

- **For \( n = 3 \)**:
\[ x = 1 + \cos\left(\frac{\pi}{6} + \pi\right) = 1 + \cos\left(\frac{7\pi}{6}\right) = 1 - \frac{\sqrt{3}}{2} \quad (\text{same as for } n=2) \]

### Step 5: Identify distinct values of \( x \)
From the calculations, we have:
1. \( x = 1 + \frac{\sqrt{3}}{2} \)
2. \( x = 1 \)
3. \( x = 1 - \frac{\sqrt{3}}{2} \)

### Conclusion
The distinct values of \( x \) satisfying the original equation are:
- \( 1 + \frac{\sqrt{3}}{2} \)
- \( 1 \)
- \( 1 - \frac{\sqrt{3}}{2} \)

Thus, the number of distinct values of \( x \) is **3**.