`cos(x+pi/3)cos(x-pi/3)=1/4 cos^2(2x)` then find the number of solution in `[-3pi,3pi]`

Answer

To solve the equation \( \cos(x + \frac{\pi}{3}) \cos(x - \frac{\pi}{3}) = \frac{1}{4} \cos^2(2x) \) and find the number of solutions in the interval \([-3\pi, 3\pi]\), we will follow these steps:

### Step 1: Use the product-to-sum identities
We can simplify the left side using the product-to-sum formula:
\[ \cos A \cos B = \frac{1}{2} (\cos(A + B) + \cos(A - B)) \]
Let \( A = x + \frac{\pi}{3} \) and \( B = x - \frac{\pi}{3} \). Then:
\[ \cos(x + \frac{\pi}{3}) \cos(x - \frac{\pi}{3}) = \frac{1}{2} \left( \cos(2x) + \cos\left(\frac{2\pi}{3}\right) \right) \]
Since \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \), we have:
\[ \cos(x + \frac{\pi}{3}) \cos(x - \frac{\pi}{3}) = \frac{1}{2} \left( \cos(2x) - \frac{1}{2} \right) \]
Thus:
\[ \cos(x + \frac{\pi}{3}) \cos(x - \frac{\pi}{3}) = \frac{1}{2} \cos(2x) - \frac{1}{4} \]

### Step 2: Set up the equation
Now we can rewrite the original equation:
\[ \frac{1}{2} \cos(2x) - \frac{1}{4} = \frac{1}{4} \cos^2(2x) \]
Multiply through by 4 to eliminate the fractions:
\[ 2 \cos(2x) - 1 = \cos^2(2x) \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \cos^2(2x) - 2 \cos(2x) + 1 = 0 \]
This can be factored as:
\[ (\cos(2x) - 1)^2 = 0 \]
Thus:
\[ \cos(2x) - 1 = 0 \implies \cos(2x) = 1 \]

### Step 4: Solve for \( x \)
The general solution for \( \cos(2x) = 1 \) is:
\[ 2x = 2n\pi \implies x = n\pi \]
where \( n \) is any integer.

### Step 5: Find the values of \( n \)
Now we need to find the integer values of \( n \) such that \( x \) lies in the interval \([-3\pi, 3\pi]\):
\[ -3\pi \leq n\pi \leq 3\pi \]
Dividing through by \( \pi \):
\[ -3 \leq n \leq 3 \]
The integer values of \( n \) that satisfy this inequality are:
\[ n = -3, -2, -1, 0, 1, 2, 3 \]

### Step 6: Count the solutions
Counting these values, we have:
- \( n = -3 \) gives \( x = -3\pi \)
- \( n = -2 \) gives \( x = -2\pi \)
- \( n = -1 \) gives \( x = -\pi \)
- \( n = 0 \) gives \( x = 0 \)
- \( n = 1 \) gives \( x = \pi \)
- \( n = 2 \) gives \( x = 2\pi \)
- \( n = 3 \) gives \( x = 3\pi \)

This gives us a total of **7 solutions** in the interval \([-3\pi, 3\pi]\).

### Final Answer
The number of solutions in the interval \([-3\pi, 3\pi]\) is **7**.