If `ln2pi ltlog_2 (2+sqrt3)lt ln3pi`, then number of roots the equation `4cos(e^x)=2^x" + 2^(-x)`

Answer

To solve the equation \( 4 \cos(e^x) = 2^x + 2^{-x} \) given the condition \( \ln(2\pi) < \log_2(2+\sqrt{3}) < \ln(3\pi) \), we will analyze the graphs of both sides of the equation and determine the number of intersections.

### Step 1: Understand the functions involved

1. **Left-hand side:** \( y_1 = 4 \cos(e^x) \)
2. **Right-hand side:** \( y_2 = 2^x + 2^{-x} \)

### Step 2: Analyze \( y_2 = 2^x + 2^{-x} \)

- The function \( y_2 = 2^x + 2^{-x} \) is symmetric about the y-axis and has a minimum value at \( x = 0 \):
\[ y_2(0) = 2^0 + 2^{-0} = 1 + 1 = 2 \]
- As \( x \to \infty \), \( y_2 \to \infty \) and as \( x \to -\infty \), \( y_2 \to \infty \) again. Thus, the graph has a minimum point at \( (0, 2) \).

### Step 3: Analyze \( y_1 = 4 \cos(e^x) \)

- The function \( y_1 \) oscillates between -4 and 4 since the cosine function oscillates between -1 and 1.
- The maximum value occurs when \( \cos(e^x) = 1 \), giving \( y_1 = 4 \).
- The function \( y_1 \) is periodic, but the argument \( e^x \) grows exponentially, causing the oscillations to become more frequent as \( x \) increases.

### Step 4: Find the maximum point of \( y_1 \)

- The maximum value of \( y_1 \) occurs when \( e^x = 2n\pi \) for \( n \in \mathbb{Z} \). Thus, the maximum points occur at:
\[ x = \ln(2n\pi) \]

### Step 5: Determine the intersections

- We need to find the number of intersections between \( y_1 \) and \( y_2 \).
- The condition \( \ln(2\pi) < \log_2(2+\sqrt{3}) < \ln(3\pi) \) implies that the maximum value of \( y_1 \) at \( x = \ln(2\pi) \) is less than the value of \( y_2 \) at \( x = \ln(2+\sqrt{3}) \) and greater than the value of \( y_2 \) at \( x = \ln(3\pi) \).

### Step 6: Analyze the values

- At \( x = \ln(2\pi) \):
\[ y_2(\ln(2\pi)) = 2^{\ln(2\pi)} + 2^{-\ln(2\pi)} = 2\pi + \frac{1}{2\pi} \]
- At \( x = \ln(3\pi) \):
\[ y_2(\ln(3\pi)) = 2^{\ln(3\pi)} + 2^{-\ln(3\pi)} = 3\pi + \frac{1}{3\pi} \]

### Step 7: Count the roots

- The oscillating nature of \( y_1 \) and the minimum of \( y_2 \) at \( y_2(0) = 2 \) indicates that there will be multiple intersections.
- Given the periodic nature of \( y_1 \) and the behavior of \( y_2 \), we can conclude that there are **6 intersections**.

### Conclusion

The number of roots of the equation \( 4 \cos(e^x) = 2^x + 2^{-x} \) is **6**.