If cos x = b.For what b do the roots of the equation form an A.P.?

Options

A. `(sqrt3)/(2)`
B. `(1)/(2)`
C. `(-1)`
D. None of these

Answer

To determine the values of \( b \) for which the roots of the equation \( \cos x = b \) form an arithmetic progression (A.P.), we can follow these steps:

### Step 1: Understand the Equation
The equation \( \cos x = b \) represents the values of \( x \) for which the cosine function equals \( b \). The cosine function oscillates between -1 and 1.

### Step 2: Analyze the Roots
The roots of \( \cos x = b \) occur at specific intervals. The general solutions for \( \cos x = b \) are given by:
\[ x = \pm \cos^{-1}(b) + 2n\pi \quad (n \in \mathbb{Z}) \]
This means there are two primary solutions in each interval of \( 2\pi \).

### Step 3: Conditions for A.P.
For the roots to form an A.P., the difference between consecutive roots must be constant. The roots can be expressed as:
\[ x_1 = \cos^{-1}(b) + 2n\pi, \quad x_2 = -\cos^{-1}(b) + 2n\pi, \quad x_3 = \cos^{-1}(b) + 2(n+1)\pi, \quad x_4 = -\cos^{-1}(b) + 2(n+1)\pi \]
To form an A.P., the difference between \( x_1 \) and \( x_2 \), and between \( x_3 \) and \( x_4 \) must be equal.

### Step 4: Evaluate Specific Cases
1. **Case \( b = 1 \)**:
- \( \cos x = 1 \) has a root at \( x = 0 + 2n\pi \).
- All roots are \( 0, 2\pi, 4\pi, \ldots \) which are in A.P.

2. **Case \( b = -1 \)**:
- \( \cos x = -1 \) has a root at \( x = \pi + 2n\pi \).
- All roots are \( \pi, 3\pi, 5\pi, \ldots \) which are in A.P.

3. **Case \( b = 0 \)**:
- \( \cos x = 0 \) has roots at \( x = \frac{\pi}{2} + n\pi \).
- All roots are \( \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \) which are in A.P.

### Conclusion
The values of \( b \) for which the roots of the equation \( \cos x = b \) form an A.P. are:
\[ b = 0, \quad b = 1, \quad b = -1 \]