The number of real solutions of the equation `"sin"e^(x)"cos" e^(x) = 2^(x-2) + 2^(-x-2)`, is

Options

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

Explanation

To find the number of real solutions of the equation

\[ \sin(e^x) \cos(e^x) = 2^{x-2} + 2^{-x-2}, \]

we can follow these steps:

### Step 1: Simplify the Left Side
Using the double angle identity for sine, we can rewrite the left side:

\[ \sin(e^x) \cos(e^x) = \frac{1}{2} \sin(2e^x). \]

### Step 2: Simplify the Right Side
The right side can be rewritten as:

\[ 2^{x-2} + 2^{-x-2} = \frac{1}{4}(2^x + 2^{-x}). \]

### Step 3: Equate Both Sides
Now, we can set the two sides equal to each other:

\[ \frac{1}{2} \sin(2e^x) = \frac{1}{4}(2^x + 2^{-x}). \]

### Step 4: Multiply Both Sides by 4
To eliminate the fractions, multiply both sides by 4:

\[ 2 \sin(2e^x) = 2^x + 2^{-x}. \]

### Step 5: Rewrite the Right Side
The right side can be expressed as:

\[ 2^x + 2^{-x} = 2 \cosh(x \ln(2)). \]

Thus, we have:

\[ 2 \sin(2e^x) = 2 \cosh(x \ln(2)). \]

### Step 6: Divide Both Sides by 2
Dividing both sides by 2 gives us:

\[ \sin(2e^x) = \cosh(x \ln(2)). \]

### Step 7: Analyze the Functions
- The sine function, \(\sin(2e^x)\), oscillates between -1 and 1 for all \(x\).
- The hyperbolic cosine function, \(\cosh(x \ln(2))\), is always greater than or equal to 1 for all \(x\).

### Step 8: Set Up the Inequality
Since \(\sin(2e^x)\) can never equal \(\cosh(x \ln(2))\) (which is always ≥ 1), we conclude that:

\[ \sin(2e^x) \leq 1 < \cosh(x \ln(2)). \]

### Step 9: Conclusion
Therefore, there are no values of \(x\) that satisfy the equation, leading us to conclude that the number of real solutions is:

\[ \boxed{0}. \]