Find the number of solution of equation 2cos x+cos 2kx =3

Answer

To find the number of solutions for the equation \(2 \cos x + \cos(2kx) = 3\), we can follow these steps:

### Step 1: Analyze the equation
The equation given is:
\[ 2 \cos x + \cos(2kx) = 3 \]
We know that the maximum value of \(\cos x\) and \(\cos(2kx)\) is 1. Therefore, the maximum value of \(2 \cos x + \cos(2kx)\) can be calculated as follows:
\[ 2 \cdot 1 + 1 = 3 \]
This means that the equation can only hold true when both \(\cos x\) and \(\cos(2kx)\) are equal to their maximum values.

### Step 2: Set conditions for solutions
For the equation to hold, we need:
\[ \cos x = 1 \quad \text{and} \quad \cos(2kx) = 1 \]
The condition \(\cos x = 1\) occurs when:
\[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
The condition \(\cos(2kx) = 1\) occurs when:
\[ 2kx = 2m\pi \quad \text{for } m \in \mathbb{Z} \]
This simplifies to:
\[ x = \frac{m\pi}{k} \]

### Step 3: Equate the two conditions
Now we have two expressions for \(x\):
1. \(x = 2n\pi\)
2. \(x = \frac{m\pi}{k}\)

To find the solutions, we set these equal to each other:
\[ 2n\pi = \frac{m\pi}{k} \]
Dividing both sides by \(\pi\) (assuming \(\pi \neq 0\)):
\[ 2n = \frac{m}{k} \]
This implies:
\[ m = 2nk \]

### Step 4: Count the number of solutions
For every integer \(n\), there corresponds an integer \(m = 2nk\). Thus, for each integer \(n\), we can find a corresponding \(m\).

Since \(n\) can take any integer value, we can conclude that there are infinitely many solutions for \(x\) in the form:
\[ x = 2n\pi \quad \text{and} \quad x = \frac{2nk\pi}{k} = \frac{m\pi}{k} \]

### Conclusion
Thus, the number of solutions to the equation \(2 \cos x + \cos(2kx) = 3\) is infinite.