Solve `cos theta * cos 2 theta * cos 3 theta = (1)/( 4), 0 le theta le pi`

Answer

To solve the equation \( \cos \theta \cdot \cos 2\theta \cdot \cos 3\theta = \frac{1}{4} \) for \( 0 \leq \theta \leq \pi \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \cos \theta \cdot \cos 2\theta \cdot \cos 3\theta = \frac{1}{4} \]
We can multiply both sides by 4 to simplify:
\[ 4 \cos \theta \cdot \cos 2\theta \cdot \cos 3\theta = 1 \]

### Step 2: Use the Product-to-Sum Formulas
We can use the product-to-sum identities to simplify the left-hand side. Recall that:
\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]
Applying this to \( \cos \theta \cdot \cos 3\theta \):
\[ 2 \cos \theta \cdot \cos 3\theta = \cos(4\theta) + \cos(2\theta) \]
Thus, we can rewrite:
\[ 4 \cos \theta \cdot \cos 2\theta \cdot \cos 3\theta = 2 \cdot (2 \cos \theta \cdot \cos 3\theta) \cdot \cos 2\theta = 2(\cos(4\theta) + \cos(2\theta)) \cdot \cos 2\theta \]

### Step 3: Set Up the New Equation
Now we have:
\[ 2(\cos(4\theta) + \cos(2\theta)) \cdot \cos 2\theta = 1 \]
This simplifies to:
\[ \cos(4\theta) + \cos(2\theta) = \frac{1}{2\cos 2\theta} \]

### Step 4: Solve for \( \cos(4\theta) \)
Using the identity \( 1 + \cos(4\theta) = 2\cos^2(2\theta) \):
\[ \cos(4\theta) = 2\cos^2(2\theta) - 1 \]
Substituting this back into the equation gives:
\[ 2\cos^2(2\theta) - 1 + \cos(2\theta) = \frac{1}{2\cos 2\theta} \]

### Step 5: Rearranging the Equation
Rearranging gives us a quadratic in \( \cos(2\theta) \):
\[ 2\cos^2(2\theta) + \cos(2\theta) - 1 + \frac{1}{2\cos 2\theta} = 0 \]
Multiplying through by \( 2\cos 2\theta \) to eliminate the fraction leads to:
\[ 4\cos^3(2\theta) + 2\cos^2(2\theta) - 2\cos 2\theta - 1 = 0 \]

### Step 6: Finding Roots
This cubic equation can be solved using numerical methods or by factoring if possible. We can find the roots for \( \cos(2\theta) \) and then solve for \( \theta \).

### Step 7: Solve for \( \theta \)
After finding the values of \( \cos(2\theta) \), we can find \( 2\theta \) and subsequently \( \theta \).

### Step 8: Check the Range
Since we need \( 0 \leq \theta \leq \pi \), we will check the values obtained from the roots to ensure they fall within this range.

### Final Values
After solving, we will find the values of \( \theta \) that satisfy the original equation.