`x_(1) and x_(2)` are two positive value of x for which ` 2 cos x,|cos x| and 3 sin^(2) x-2` are in GP. The minimum value of ` |x_(1)-x_(2)|` is equal to

Options

A. `(4pi)/(3)`
B. `(pi)/6`
C. `2 cos^(-1)""((2)/(3))`
D. `cos^(-1)""((2)/(3))`

Answer

To solve the problem, we need to find the minimum value of \(|x_1 - x_2|\) given that \(2 \cos x\), \(|\cos x|\), and \(3 \sin^2 x - 2\) are in geometric progression (GP).

### Step-by-Step Solution

1. **Understanding the GP Condition**:
For three numbers \(a\), \(b\), and \(c\) to be in GP, the condition is:
\[ b^2 = ac \]
Here, let \(a = 2 \cos x\), \(b = |\cos x|\), and \(c = 3 \sin^2 x - 2\).

2. **Setting Up the Equation**:
We can write the GP condition as:
\[ |\cos x|^2 = (2 \cos x)(3 \sin^2 x - 2) \]

3. **Expressing \(\sin^2 x\)**:
Recall that \(\sin^2 x = 1 - \cos^2 x\). Substituting this into the equation gives:
\[ |\cos x|^2 = (2 \cos x)(3(1 - \cos^2 x) - 2) \]

4. **Simplifying the Equation**:
Expanding the right-hand side:
\[ |\cos x|^2 = 2 \cos x (3 - 3 \cos^2 x - 2) = 2 \cos x (1 - 3 \cos^2 x) \]
This simplifies to:
\[ |\cos x|^2 = 2 \cos x - 6 \cos^3 x \]

5. **Considering Cases for \(|\cos x|\)**:
Since \(|\cos x| = \cos x\) when \(\cos x \geq 0\) and \(|\cos x| = -\cos x\) when \(\cos x < 0\), we will consider the case where \(\cos x \geq 0\) first.

6. **Rearranging the Equation**:
Rearranging gives:
\[ 6 \cos^3 x - 3 \cos x + 2 = 0 \]

7. **Factoring the Cubic Equation**:
We can factor this cubic equation:
\[ (3 \cos x + 2)(2 \cos^2 x - 1) = 0 \]
This gives us two cases:
- \(3 \cos x + 2 = 0 \Rightarrow \cos x = -\frac{2}{3}\) (not valid for positive \(x\))
- \(2 \cos^2 x - 1 = 0 \Rightarrow \cos^2 x = \frac{1}{2} \Rightarrow \cos x = \frac{1}{\sqrt{2}} = \frac{1}{2}\) (valid)

8. **Finding Values of \(x\)**:
From \(\cos x = \frac{1}{2}\), we have:
\[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
The positive solutions are:
\[ x_1 = \frac{\pi}{3}, \quad x_2 = \frac{5\pi}{3} \]

9. **Calculating \(|x_1 - x_2|\)**:
The minimum value of \(|x_1 - x_2|\) is:
\[ |x_1 - x_2| = \left| \frac{\pi}{3} - \frac{5\pi}{3} \right| = \left| -\frac{4\pi}{3} \right| = \frac{4\pi}{3} \]

10. **Final Answer**:
The minimum value of \(|x_1 - x_2|\) is:
\[ \frac{\pi}{6} \]