The number of solutions of the equation `cos (x+pi/3)cos (pi/3-x) = 1/4 cos^(2) 2x, x in [-3pi,3pi]` is :

Options

A. 8
B. 5
C. 6
D. 7

Answer

To solve the equation \( \cos(x + \frac{\pi}{3}) \cos(\frac{\pi}{3} - x) = \frac{1}{4} \cos^2(2x) \) for \( x \) in the interval \( [-3\pi, 3\pi] \), we can follow these steps:

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

### Step 2: Set Up the Equation
Now we set the left-hand side equal to the right-hand side:
\[ \frac{1}{2} \cos(2x) - \frac{1}{4} = \frac{1}{4} \cos^2(2x) \]
Multiply through by 4 to eliminate the fractions:
\[ 2 \cos(2x) - 1 = \cos^2(2x) \]

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

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

### Step 5: Find Solutions in the Interval
We need to find integer values of \( n \) such that:
\[ -3\pi \leq n\pi \leq 3\pi \]
Dividing through by \( \pi \):
\[ -3 \leq n \leq 3 \]
The integer values of \( n \) that satisfy this inequality are:
\[ n = -3, -2, -1, 0, 1, 2, 3 \]
This gives us a total of 7 solutions.

### Conclusion
The number of solutions of the equation \( \cos(x + \frac{\pi}{3}) \cos(\frac{\pi}{3} - x) = \frac{1}{4} \cos^2(2x) \) in the interval \( [-3\pi, 3\pi] \) is \( \boxed{7} \).