General solution of the equation,
`cos x cdot cos 6x = -1` is =

Options

A. has 50 solutions in ` [0, 100 pi]`
B. has 3 solutions in ` [ 0, 3 pi] `
C. has even number of solutions in ` ( 3 pi , 13 pi ) `
D. has one solution in ` [ (pi)/(2) , pi] `

Answer

To find the general solution of the equation \( \cos x \cdot \cos 6x = -1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos x \cdot \cos 6x = -1 \]

### Step 2: Multiply both sides by 2
To use the product-to-sum identities, we multiply both sides by 2:
\[ 2 \cos x \cdot \cos 6x = -2 \]

### Step 3: Apply the product-to-sum identity
Using the identity \( 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \), we can rewrite the left side:
\[ \cos(x + 6x) + \cos(x - 6x) = -2 \]
This simplifies to:
\[ \cos(7x) + \cos(-5x) = -2 \]

### Step 4: Simplify the cosine term
Since \( \cos(-\theta) = \cos(\theta) \), we can rewrite the equation as:
\[ \cos(7x) + \cos(5x) = -2 \]

### Step 5: Analyze the equation
The maximum value of \( \cos \) is 1. Therefore, for the sum of two cosines to equal -2, both must be at their minimum value:
\[ \cos(7x) = -1 \quad \text{and} \quad \cos(5x) = -1 \]

### Step 6: Find the values of \( x \)
The general solutions for \( \cos \theta = -1 \) occur at:
\[ \theta = (2n + 1) \pi \quad \text{for } n \in \mathbb{Z} \]
Thus, we have:
1. For \( \cos(7x) = -1 \):
\[ 7x = (2n + 1) \pi \implies x = \frac{(2n + 1) \pi}{7} \]

2. For \( \cos(5x) = -1 \):
\[ 5x = (2m + 1) \pi \implies x = \frac{(2m + 1) \pi}{5} \]

### Step 7: Combine the solutions
The general solution can be expressed as:
\[ x = \frac{(2n + 1) \pi}{7} \quad \text{and} \quad x = \frac{(2m + 1) \pi}{5} \]
where \( n, m \in \mathbb{Z} \).

### Final General Solution
The general solution of the equation \( \cos x \cdot \cos 6x = -1 \) is:
\[ x = \frac{(2n + 1) \pi}{7} \quad \text{and} \quad x = \frac{(2m + 1) \pi}{5}, \quad n, m \in \mathbb{Z} \]

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