The number of values of x lying in the inteval `-(2pi, 2pi)` satisfying the equation `1+cos 10x cos 6x=2 cos^(2)8x+sin^(2)8x` is equal to

Answer

To solve the equation \(1 + \cos(10x) \cos(6x) = 2 \cos^2(8x) + \sin^2(8x)\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 1 + \cos(10x) \cos(6x) = 2 \cos^2(8x) + \sin^2(8x) \]

### Step 2: Simplify the right-hand side
Using the identity \(\sin^2(a) = 1 - \cos^2(a)\), we can rewrite the right-hand side:
\[ 2 \cos^2(8x) + \sin^2(8x) = 2 \cos^2(8x) + (1 - \cos^2(8x)) = \cos^2(8x) + 1 \]
Thus, the equation becomes:
\[ 1 + \cos(10x) \cos(6x) = \cos^2(8x) + 1 \]

### Step 3: Cancel out the 1s
Subtracting 1 from both sides gives:
\[ \cos(10x) \cos(6x) = \cos^2(8x) \]

### Step 4: Use the product-to-sum identities
We can use the product-to-sum identity:
\[ \cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \]
Applying this to \(\cos(10x) \cos(6x)\):
\[ \cos(10x) \cos(6x) = \frac{1}{2} [\cos(16x) + \cos(4x)] \]
Thus, our equation now becomes:
\[ \frac{1}{2} [\cos(16x) + \cos(4x)] = \cos^2(8x) \]

### Step 5: Multiply through by 2
Multiplying both sides by 2 results in:
\[ \cos(16x) + \cos(4x) = 2 \cos^2(8x) \]

### Step 6: Rewrite the right-hand side
Using the identity \(2 \cos^2(a) = 1 + \cos(2a)\), we can rewrite the right-hand side:
\[ \cos(16x) + \cos(4x) = 1 + \cos(16x) \]
This simplifies to:
\[ \cos(4x) = 1 \]

### Step 7: Solve for \(x\)
The equation \(\cos(4x) = 1\) implies:
\[ 4x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ x = \frac{n\pi}{2} \]

### Step 8: Find the values of \(x\) in the interval \(-2\pi < x < 2\pi\)
Now we need to find the integer values of \(n\) such that:
\[ -2\pi < \frac{n\pi}{2} < 2\pi \]
Multiplying through by 2:
\[ -4 < n < 4 \]
This gives possible integer values for \(n\) as:
\[ n = -3, -2, -1, 0, 1, 2, 3 \]
This results in the following values of \(x\):
- For \(n = -3\): \(x = -\frac{3\pi}{2}\)
- For \(n = -2\): \(x = -\pi\)
- For \(n = -1\): \(x = -\frac{\pi}{2}\)
- For \(n = 0\): \(x = 0\)
- For \(n = 1\): \(x = \frac{\pi}{2}\)
- For \(n = 2\): \(x = \pi\)
- For \(n = 3\): \(x = \frac{3\pi}{2}\)

### Step 9: Count the solutions
The values of \(x\) are:
\(-\frac{3\pi}{2}, -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\)

Thus, there are a total of **7 values** of \(x\) in the interval \(-2\pi < x < 2\pi\).

### Final Answer
The number of values of \(x\) lying in the interval \(-2\pi < x < 2\pi\) satisfying the equation is \(7\).

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