Solve: `costheta cos2theta cos3theta =(1)/(4)`

Answer

To solve the equation \( \cos \theta \cos 2\theta \cos 3\theta = \frac{1}{4} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos \theta \cos 2\theta \cos 3\theta = \frac{1}{4} \]

### Step 2: Multiply both sides by 2
To simplify the expression, we can multiply both sides by 2:
\[ 2 \cos \theta \cos 2\theta \cos 3\theta = \frac{1}{2} \]

### Step 3: Use the product-to-sum identities
We can use the identity \( 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \). Let’s apply this to \( \cos \theta \) and \( \cos 3\theta \):
\[ 2 \cos \theta \cos 3\theta = \cos(4\theta) + \cos(2\theta) \]
Thus, we can rewrite the equation as:
\[ \cos(4\theta) + \cos(2\theta) \cos 2\theta = \frac{1}{2} \]

### Step 4: Rearranging the equation
Now we can rearrange the equation:
\[ \cos(4\theta) + \frac{1}{2} \cos^2(2\theta) = \frac{1}{2} \]
This simplifies to:
\[ \cos(4\theta) + \frac{1}{2} \cos^2(2\theta) - \frac{1}{2} = 0 \]
or
\[ \cos(4\theta) + \frac{1}{2} \cos^2(2\theta) = 0 \]

### Step 5: Factor the equation
We can factor out \( \cos(4\theta) \):
\[ \cos(4\theta) \left( \cos(2\theta) + \frac{1}{2} \right) = 0 \]

### Step 6: Solve for \( \cos(4\theta) = 0 \)
Setting \( \cos(4\theta) = 0 \):
\[ 4\theta = \frac{(2n + 1)\pi}{2} \quad \text{for } n \in \mathbb{Z} \]
This gives:
\[ \theta = \frac{(2n + 1)\pi}{8} \]

### Step 7: Solve for \( \cos(2\theta) + \frac{1}{2} = 0 \)
Setting \( \cos(2\theta) + \frac{1}{2} = 0 \):
\[ \cos(2\theta) = -\frac{1}{2} \]
This occurs at:
\[ 2\theta = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad 2\theta = \frac{4\pi}{3} + 2n\pi \]
Thus, we have:
\[ \theta = \frac{\pi}{3} + n\pi \quad \text{or} \quad \theta = \frac{2\pi}{3} + n\pi \]

### Final Solutions
Combining both sets of solutions, we have:
1. \( \theta = \frac{(2n + 1)\pi}{8} \)
2. \( \theta = \frac{\pi}{3} + n\pi \)
3. \( \theta = \frac{2\pi}{3} + n\pi \)