If the expression exp `{1+|cosx|+cos^(2)x+|cos^(3)x|+ . . . . oo)log_(e)4}` satisfies the equation `y^(2)-20y+64=0" for "0ltxltpi`, then the set of value of x is

Options

A. `{pi//3,2pi//3}`
B. `{pi//2,pi//2}`
C. `{pi//2,0,2pi//3}`
D. `{pi//3,pi//2,2pi//3}`

Explanation

To solve the problem step by step, we will analyze the given expression and the quadratic equation separately.

### Step 1: Analyze the Infinite Series
The expression given is:
\[ \exp \left( 1 + |\cos x| + \cos^2 x + |\cos^3 x| + \ldots \right) \log_e 4 \]

The series inside the exponential can be expressed as:
\[ S = 1 + |\cos x| + \cos^2 x + |\cos^3 x| + \ldots \]

This series can be split into two parts based on whether \( \cos x \) is positive or negative. However, since we are taking the absolute value, we can treat it as a geometric series.

### Step 2: Identify the Geometric Series
The series can be rewritten as:
\[ S = 1 + |\cos x| + |\cos x|^2 + |\cos x|^3 + \ldots \]

This is a geometric series with the first term \( a = 1 \) and common ratio \( r = |\cos x| \).

### Step 3: Sum of the Geometric Series
The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{1}{1 - |\cos x|} \]
for \( |r| < 1 \). Therefore,
\[ S = \frac{1}{1 - |\cos x|} \]

### Step 4: Substitute Back into the Exponential
Now substituting \( S \) back into the exponential, we have:
\[ \exp\left( \frac{1}{1 - |\cos x|} \right) \log_e 4 \]

### Step 5: Set Up the Quadratic Equation
The problem states that this expression satisfies the equation:
\[ y^2 - 20y + 64 = 0 \]

### Step 6: Solve the Quadratic Equation
To solve the quadratic equation, we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -20, c = 64 \):
\[ y = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \]
\[ y = \frac{20 \pm \sqrt{400 - 256}}{2} \]
\[ y = \frac{20 \pm \sqrt{144}}{2} \]
\[ y = \frac{20 \pm 12}{2} \]
Thus, the solutions are:
\[ y = \frac{32}{2} = 16 \quad \text{and} \quad y = \frac{8}{2} = 4 \]

### Step 7: Set Up the Exponential Equations
Now we have two cases:
1. \( \exp\left( \frac{1}{1 - |\cos x|} \right) \log_e 4 = 4 \)
2. \( \exp\left( \frac{1}{1 - |\cos x|} \right) \log_e 4 = 16 \)

### Step 8: Solve Each Case
**Case 1:**
\[ \exp\left( \frac{1}{1 - |\cos x|} \right) = 1 \]
This implies:
\[ \frac{1}{1 - |\cos x|} = 0 \Rightarrow 1 - |\cos x| \to \infty \text{ (not possible)} \]

**Case 2:**
\[ \exp\left( \frac{1}{1 - |\cos x|} \right) = 4 \]
This implies:
\[ \frac{1}{1 - |\cos x|} = \log_e 4 \]
Thus,
\[ 1 - |\cos x| = \frac{1}{\log_e 4} \]
So,
\[ |\cos x| = 1 - \frac{1}{\log_e 4} \]

### Step 9: Find Values of \( x \)
Now we can find \( x \) such that:
1. \( |\cos x| = 0 \) gives \( x = \frac{\pi}{2} \)
2. \( |\cos x| = \frac{1}{2} \) gives \( x = \frac{\pi}{3}, \frac{5\pi}{3} \) (but only \( \frac{5\pi}{3} \) is valid in \( (0, \pi) \))

Thus, the possible values of \( x \) are:
- \( x = \frac{\pi}{2} \)
- \( x = \frac{\pi}{3} \)
- \( x = \frac{2\pi}{3} \)

### Final Answer
The set of values of \( x \) is:
\[ \left\{ \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3} \right\} \]