If `|{:(1+cos^(2)theta,sin^(2)theta,4cos6theta),(cos^(2)theta,1+sin^(2)theta,4cos6theta),(cos^(2)theta,sin^(2)theta,1+4cos6theta):}|=0`, and `theta in (0,(pi)/(3))`, then value of `theta` is

Options

A. `(7pi)/(36)`
B. `(7pi)/(24)`
C. `(pi)/(9)`
D. `(pi)/(4)`

Answer

To solve the problem, we need to evaluate the determinant given and find the value of \( \theta \) in the interval \( (0, \frac{\pi}{3}) \) such that the determinant equals zero.

### Step-by-step Solution:

1. **Write the Determinant**:
We have the determinant:
\[ D = \begin{vmatrix} 1 + \cos^2 \theta & \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]
We need to find when \( |D| = 0 \).

2. **Perform Column Operations**:
We can simplify the determinant by performing column operations. Let's add the first column to the second column:
\[ D = \begin{vmatrix} 1 + \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]
This results in:
\[ D = \begin{vmatrix} 1 + \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]

3. **Subtract Rows**:
Now, subtract the second row from the first row and the third row from the second row:
\[ D = \begin{vmatrix} 1 + \cos^2 \theta - \cos^2 \theta & 1 + \sin^2 \theta - (1 + \sin^2 \theta) & 4 \cos 6\theta - 4 \cos 6\theta \\ \cos^2 \theta - \cos^2 \theta & 1 + \sin^2 \theta - \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]

4. **Expand the Determinant**:
Now, we can expand the determinant:
\[ D = 1 \cdot \begin{vmatrix} 1 + \sin^2 \theta & 4 \cos 6\theta \\ \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]
This gives:
\[ D = (1 + \sin^2 \theta)(1 + 4 \cos 6\theta) - (4 \cos 6\theta \sin^2 \theta) \]

5. **Set the Determinant to Zero**:
Set \( D = 0 \):
\[ (1 + \sin^2 \theta)(1 + 4 \cos 6\theta) - 4 \cos 6\theta \sin^2 \theta = 0 \]
Simplifying this leads to:
\[ 1 + \sin^2 \theta + 4 \cos 6\theta + 4 \sin^2 \theta \cos 6\theta - 4 \cos 6\theta \sin^2 \theta = 0 \]
Rearranging gives:
\[ 1 + 5 \sin^2 \theta + 4 \cos 6\theta(1 - \sin^2 \theta) = 0 \]

6. **Solve for \( \theta \)**:
We can isolate \( \cos 6\theta \):
\[ 4 \cos 6\theta = - (1 + 5 \sin^2 \theta) \]
This leads to:
\[ \cos 6\theta = -\frac{1 + 5 \sin^2 \theta}{4} \]
Since \( \cos 6\theta \) must be between -1 and 1, we can find suitable values of \( \theta \).

7. **Find \( \theta \)**:
The equation \( \cos 6\theta = -\frac{1}{2} \) gives:
\[ 6\theta = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad 6\theta = \frac{4\pi}{3} + 2k\pi \]
Solving for \( \theta \):
\[ \theta = \frac{\pi}{9} \quad \text{(for } k = 0\text{)} \]
Since \( \theta \) must be in \( (0, \frac{\pi}{3}) \), we take \( \theta = \frac{\pi}{9} \).

### Final Answer:
The value of \( \theta \) is \( \frac{\pi}{9} \).