The general solution of `cos x cos 6x =-1` is

Options

A. `x=(2n+1) pi, n in Z`
B. `x=2n pi, n in Z`
C. `x= n pi, n in Z`
D. none of these

Explanation

To solve the equation \( \cos x \cos 6x = -1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos x \cos 6x = -1 \]
To simplify, we can multiply both sides by 2:
\[ 2 \cos x \cos 6x = -2 \]

### Step 2: Use the product-to-sum identities
Using the product-to-sum identities, we know that:
\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]
Applying this to our equation:
\[ \cos(x + 6x) + \cos(x - 6x) = -2 \]
This simplifies to:
\[ \cos(7x) + \cos(5x) = -2 \]

### Step 3: Analyze the cosine function
The cosine function has a maximum value of 1 and a minimum value of -1. Therefore, for the sum \( \cos(7x) + \cos(5x) \) to equal -2, both \( \cos(7x) \) and \( \cos(5x) \) must be equal to -1:
\[ \cos(7x) = -1 \quad \text{and} \quad \cos(5x) = -1 \]

### Step 4: Solve for \( x \)
The general solution for \( \cos \theta = -1 \) is given by:
\[ \theta = (2n + 1)\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, we have:
1. For \( \cos(7x) = -1 \):
\[ 7x = (2n_1 + 1)\pi \quad \Rightarrow \quad x = \frac{(2n_1 + 1)\pi}{7} \]
2. For \( \cos(5x) = -1 \):
\[ 5x = (2n_2 + 1)\pi \quad \Rightarrow \quad x = \frac{(2n_2 + 1)\pi}{5} \]

### Step 5: Set the equations equal
Since both expressions equal \( x \), we can set them equal to each other:
\[ \frac{(2n_1 + 1)\pi}{7} = \frac{(2n_2 + 1)\pi}{5} \]
Eliminating \( \pi \) from both sides gives:
\[ \frac{2n_1 + 1}{7} = \frac{2n_2 + 1}{5} \]
Cross-multiplying yields:
\[ 5(2n_1 + 1) = 7(2n_2 + 1) \]
Expanding this gives:
\[ 10n_1 + 5 = 14n_2 + 7 \]
Rearranging gives:
\[ 10n_1 - 14n_2 = 2 \]
This can be simplified to:
\[ 5n_1 - 7n_2 = 1 \]

### Step 6: General solution
The general solution for \( n_1 \) and \( n_2 \) can be expressed in terms of integers. The integer solutions can be found, but the simplest form of the solution for \( x \) can be expressed as:
\[ x = (2n + 1)\frac{\pi}{7} \quad \text{or} \quad x = (2n + 1)\frac{\pi}{5} \]
This leads us to conclude that the general solution for \( x \) can be expressed as:
\[ x = (2n + 1)\frac{\pi}{7} \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
Thus, the general solution of \( \cos x \cos 6x = -1 \) is:
\[ x = (2n + 1)\frac{\pi}{7} \quad \text{where } n \in \mathbb{Z} \]