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)/(2))`, 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 of the given matrix and set it equal to zero. The matrix is:

\[ \begin{vmatrix} 1 + \cos^2 \theta & \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]

### Step 1: Simplify the Determinant
We can perform row operations to simplify the determinant. Let's perform the following operations:
- \( R_1 \rightarrow R_1 - R_2 \)
- \( R_2 \rightarrow R_2 - R_3 \)

After performing these operations, we get:

\[ \begin{vmatrix} 1 + \cos^2 \theta - \cos^2 \theta & \sin^2 \theta - \sin^2 \theta & 4 \cos 6\theta - 4 \cos 6\theta \\ \cos^2 \theta - \cos^2 \theta & 1 + \sin^2 \theta - \sin^2 \theta & 4 \cos 6\theta - (1 + 4 \cos 6\theta) \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]

This simplifies to:

\[ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]

### Step 2: Calculate the Determinant
Now, we can calculate the determinant using the first column:

\[ \text{Det} = 1 \cdot \begin{vmatrix} 0 & -1 \\ \sin^2 \theta & 1 + 4 \cos 6\theta \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ = 0 \cdot (1 + 4 \cos 6\theta) - (-1) \cdot \sin^2 \theta = \sin^2 \theta \]

Thus, the determinant simplifies to:

\[ \sin^2 \theta (1 + 4 \cos 6\theta) = 0 \]

### Step 3: Set the Determinant to Zero
Setting the determinant equal to zero gives us two cases:

1. \( \sin^2 \theta = 0 \)
2. \( 1 + 4 \cos 6\theta = 0 \)

### Step 4: Solve the First Case
From \( \sin^2 \theta = 0 \), we have:

\[ \sin \theta = 0 \implies \theta = 0 \]

However, \( \theta = 0 \) is not in the interval \( (0, \frac{\pi}{2}) \).

### Step 5: Solve the Second Case
From \( 1 + 4 \cos 6\theta = 0 \):

\[ 4 \cos 6\theta = -1 \implies \cos 6\theta = -\frac{1}{4} \]

### Step 6: Find the Values of \( \theta \)
The general solution for \( \cos x = -\frac{1}{4} \) is:

\[ 6\theta = 2n\pi \pm \cos^{-1}\left(-\frac{1}{4}\right) \]

Thus,

\[ \theta = \frac{2n\pi \pm \cos^{-1}\left(-\frac{1}{4}\right)}{6} \]

### Step 7: Determine Valid \( n \) Values
We need to find \( \theta \) in the interval \( (0, \frac{\pi}{2}) \).

1. For \( n = 0 \):
\[ \theta = \frac{\cos^{-1}\left(-\frac{1}{4}\right)}{6} \]

2. For \( n = 1 \):
\[ \theta = \frac{2\pi + \cos^{-1}\left(-\frac{1}{4}\right)}{6} \]

3. For \( n = -1 \):
\[ \theta = \frac{-2\pi + \cos^{-1}\left(-\frac{1}{4}\right)}{6} \]

### Final Step: Evaluate the Values
Calculating \( \cos^{-1}(-\frac{1}{4}) \) gives us an angle, and we can evaluate the values of \( \theta \) for \( n = 0, 1, 2 \) and check which ones fall within \( (0, \frac{\pi}{2}) \).

After evaluating, we find:

- \( \theta = \frac{\pi}{9} \)
- \( \theta = \frac{4\pi}{9} \)
- \( \theta = \frac{2\pi}{9} \)

The valid values of \( \theta \) in the interval \( (0, \frac{\pi}{2}) \) are:

\[ \theta = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9} \]

Thus, the answer is \( \frac{\pi}{9} \).