The number of real roots of the equation `(sin 2^(x)) cos (2^(x))=(1)/(4) (2^(x)+2^(-x))` is equal to

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the equation \( \sin(2^x) \cos(2^x) = \frac{1}{4} (2^x + 2^{-x}) \), we will follow these steps:

### Step 1: Rewrite the Left-Hand Side
Using the double angle identity for sine, we can rewrite the left-hand side:
\[ \sin(2^x) \cos(2^x) = \frac{1}{2} \sin(2 \cdot 2^x) = \frac{1}{2} \sin(2^{x+1}) \]
Thus, the equation becomes:
\[ \frac{1}{2} \sin(2^{x+1}) = \frac{1}{4} (2^x + 2^{-x}) \]

### Step 2: Simplify the Equation
Multiplying both sides by 4 to eliminate the fractions gives:
\[ 2 \sin(2^{x+1}) = 2^x + 2^{-x} \]

### Step 3: Rewrite the Right-Hand Side
The right-hand side can be rewritten using the property of exponents:
\[ 2^x + 2^{-x} = 2^x + \frac{1}{2^x} = 2^x + 2^{-x} \]
This expression is always greater than or equal to 2 (by the AM-GM inequality), since:
\[ \frac{2^x + 2^{-x}}{2} \geq \sqrt{2^x \cdot 2^{-x}} = 1 \]
Thus, \( 2^x + 2^{-x} \geq 2 \).

### Step 4: Analyze the Left-Hand Side
The left-hand side, \( 2 \sin(2^{x+1}) \), oscillates between -2 and 2, since the sine function has a range of [-1, 1].

### Step 5: Set Up the Inequality
From the previous steps, we know:
\[ 2 \sin(2^{x+1}) \leq 2 \]
And from the AM-GM inequality:
\[ 2^x + 2^{-x} \geq 2 \]
Thus, we have:
\[ 2 \sin(2^{x+1}) = 2^x + 2^{-x} \geq 2 \]

### Step 6: Determine Possible Values
Since \( 2 \sin(2^{x+1}) \) can only equal \( 2^x + 2^{-x} \) when both sides are equal to 2, we need to find when:
\[ 2 \sin(2^{x+1}) = 2 \]
This implies:
\[ \sin(2^{x+1}) = 1 \]
The sine function equals 1 at \( 2^{x+1} = \frac{\pi}{2} + 2k\pi \) for \( k \in \mathbb{Z} \).

### Step 7: Solve for x
This leads to:
\[ 2^{x+1} = \frac{\pi}{2} + 2k\pi \]
Taking logarithm base 2:
\[ x + 1 = \log_2\left(\frac{\pi}{2} + 2k\pi\right) \]
Thus:
\[ x = \log_2\left(\frac{\pi}{2} + 2k\pi\right) - 1 \]

### Step 8: Count the Real Roots
The expression \( \frac{\pi}{2} + 2k\pi \) is positive for all integers \( k \), and since \( k \) can take any integer value, there are infinitely many solutions for \( x \).

### Final Answer
The number of real roots of the equation is infinite.

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