The solution of `costheta.cos2theta.cos3theta=1/4,0 lt theta lt pi/4` is:

Options

A. `pi/8`
B. `pi/6`
C. `pi/9`
D. `pi/12`

Answer

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

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

### Step 2: Use the product-to-sum identities
We can use the identity \( 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \). Here, let \( A = \theta \) and \( B = 3\theta \):
\[ 2 \cos \theta \cdot \cos 3\theta = \cos(4\theta) + \cos(2\theta) \]
Thus, we can rewrite our equation as:
\[ 2 \cos \theta \cdot \cos 3\theta = \frac{1 - \cos 2\theta}{2} \]
Substituting this back into our equation gives:
\[ \cos(4\theta) + \cos(2\theta) = 1 \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ \cos(4\theta) + \cos(2\theta) - 1 = 0 \]

### Step 4: Factor the equation
We can factor this equation:
\[ \cos(4\theta) + \cos(2\theta) - 1 = 0 \]
This can be factored as:
\[ \cos(4\theta) = 1 - \cos(2\theta) \]

### Step 5: Solve for \( \cos(4\theta) \)
We know that \( \cos(4\theta) = 0 \) or \( 2\cos(2\theta) + 1 = 0 \).

1. **Case 1:** \( \cos(4\theta) = 0 \)
\[ 4\theta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
\[ \theta = \frac{\pi}{8} + \frac{n\pi}{4} \]

2. **Case 2:** \( 2\cos(2\theta) + 1 = 0 \)
\[ \cos(2\theta) = -\frac{1}{2} \]
\[ 2\theta = \frac{2\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]
\[ \theta = \frac{\pi}{3} + n\pi \]

### Step 6: Finding valid solutions
Now we need to find valid solutions in the interval \( 0 < \theta < \frac{\pi}{4} \).

- From **Case 1:** \( \theta = \frac{\pi}{8} \) is valid.
- From **Case 2:** \( \theta = \frac{\pi}{3} \) is not valid since \( \frac{\pi}{3} > \frac{\pi}{4} \).

### Conclusion
The only valid solution in the interval \( 0 < \theta < \frac{\pi}{4} \) is:
\[ \theta = \frac{\pi}{8} \]