Find all the values of ` theta `satisfying the equation ` cos 2 theta - cos 8 theta + cos 6 theta =1,` such that `0 le theta le pi.`

Answer

To solve the equation \( \cos 2\theta - \cos 8\theta + \cos 6\theta = 1 \) for \( 0 \leq \theta \leq \pi \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation:
\[ \cos 2\theta + \cos 6\theta - \cos 8\theta = 1 \]

### Step 2: Using the Cosine Addition Formula
We can use the cosine addition formula:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
Let \( A = 2\theta \) and \( B = 6\theta \). Then,
\[ \cos 2\theta + \cos 6\theta = 2 \cos\left(\frac{2\theta + 6\theta}{2}\right) \cos\left(\frac{2\theta - 6\theta}{2}\right) = 2 \cos(4\theta) \cos(-2\theta) \]
Since \( \cos(-x) = \cos x \), we have:
\[ \cos 2\theta + \cos 6\theta = 2 \cos(4\theta) \cos(2\theta) \]

### Step 3: Substituting Back into the Equation
Substituting back into the equation gives:
\[ 2 \cos(4\theta) \cos(2\theta) - \cos 8\theta = 1 \]

### Step 4: Rewriting \( \cos 8\theta \)
Using the double angle formula, we can express \( \cos 8\theta \) as:
\[ \cos 8\theta = 2 \cos^2(4\theta) - 1 \]
So, the equation becomes:
\[ 2 \cos(4\theta) \cos(2\theta) - (2 \cos^2(4\theta) - 1) = 1 \]

### Step 5: Simplifying the Equation
Simplifying gives:
\[ 2 \cos(4\theta) \cos(2\theta) - 2 \cos^2(4\theta) + 1 = 1 \]
This simplifies to:
\[ 2 \cos(4\theta) \cos(2\theta) - 2 \cos^2(4\theta) = 0 \]

### Step 6: Factoring Out Common Terms
Factoring out \( 2 \cos(4\theta) \):
\[ 2 \cos(4\theta) (\cos(2\theta) - \cos(4\theta)) = 0 \]
This gives us two cases to solve:
1. \( \cos(4\theta) = 0 \)
2. \( \cos(2\theta) - \cos(4\theta) = 0 \)

### Step 7: Solving \( \cos(4\theta) = 0 \)
The solutions for \( \cos(4\theta) = 0 \) are:
\[ 4\theta = \frac{(2n + 1)\pi}{2} \implies \theta = \frac{(2n + 1)\pi}{8} \]
For \( n = 0, 1, 2, 3 \):
- \( n = 0 \): \( \theta = \frac{\pi}{8} \)
- \( n = 1 \): \( \theta = \frac{3\pi}{8} \)
- \( n = 2 \): \( \theta = \frac{5\pi}{8} \)
- \( n = 3 \): \( \theta = \frac{7\pi}{8} \)
- \( n = 4 \): \( \theta = \frac{9\pi}{8} \) (not valid since \( > \pi \))

### Step 8: Solving \( \cos(2\theta) - \cos(4\theta) = 0 \)
This implies:
\[ \cos(2\theta) = \cos(4\theta) \]
Using the identity \( \cos A = \cos B \) gives:
\[ 2\theta = 4\theta + 2k\pi \quad \text{or} \quad 2\theta = -4\theta + 2k\pi \]
From \( 2\theta = 4\theta + 2k\pi \):
\[ -2\theta = 2k\pi \implies \theta = -k\pi \quad \text{(not valid)} \]
From \( 2\theta = -4\theta + 2k\pi \):
\[ 6\theta = 2k\pi \implies \theta = \frac{k\pi}{3} \]
For \( k = 0, 1, 2, 3 \):
- \( k = 0 \): \( \theta = 0 \)
- \( k = 1 \): \( \theta = \frac{\pi}{3} \)
- \( k = 2 \): \( \theta = \frac{2\pi}{3} \)
- \( k = 3 \): \( \theta = \pi \)

### Step 9: Collecting All Valid Solutions
Combining all the valid solutions:
- From \( \cos(4\theta) = 0 \): \( \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \)
- From \( \cos(2\theta) - \cos(4\theta) = 0 \): \( 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi \)

### Final Values of \( \theta \)
Thus, the final values of \( \theta \) satisfying the equation in the interval \( [0, \pi] \) are:
\[ 0, \frac{\pi}{8}, \frac{3\pi}{8}, \frac{\pi}{3}, \frac{5\pi}{8}, \frac{2\pi}{3}, \frac{7\pi}{8}, \pi \]