The values of x in `(0, pi)` satisfying the equation.
`|{:(1+"sin"^(2)x, "sin"^(2)x, "sin"^(2)x), ("cos"^(2)x, 1+"cos"^(2)x, "cos"^(2)x), (4"sin" 2x, 4"sin"2x, 1+4"sin" 2x):}| = 0`, are

Options

A. `(pi)/(12), (7pi)/(12)`
B. `(5pi)/(12), (7pi)/(12)`
C. `(7pi)/(12), (11pi)/(12)`
D. `(pi)/(12), (11pi)/(12)`

Explanation

To solve the equation given by the determinant of the matrix \( T \) in the interval \( (0, \pi) \), we will follow these steps:

### Step 1: Write down the matrix and set up the determinant
The matrix \( T \) is given as:
\[ T = \begin{pmatrix} 1 + \sin^2 x & \sin^2 x & \sin^2 x \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4 \sin 2x & 4 \sin 2x & 1 + 4 \sin 2x \end{pmatrix} \]
We need to find the values of \( x \) such that \( \det(T) = 0 \).

### Step 2: Apply column transformations
We will perform column operations to simplify the determinant calculation.

1. **Transform Column 1**: \( C_1 \leftarrow C_1 - C_2 \)
\[ C_1 = \begin{pmatrix} (1 + \sin^2 x - \sin^2 x) \\ (\cos^2 x - (1 + \cos^2 x)) \\ (4 \sin 2x - 4 \sin 2x) \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \]

2. **Transform Column 2**: \( C_2 \leftarrow C_2 - C_3 \)
\[ C_2 = \begin{pmatrix} \sin^2 x - \sin^2 x \\ (1 + \cos^2 x - \cos^2 x) \\ (4 \sin 2x - (1 + 4 \sin 2x)) \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \]

The new matrix becomes:
\[ \begin{pmatrix} 1 & 0 & \sin^2 x \\ -1 & 1 & \cos^2 x \\ 0 & -1 & 1 + 4 \sin 2x \end{pmatrix} \]

### Step 3: Calculate the determinant
Now we can calculate the determinant:
\[ \det(T) = 1 \cdot \begin{vmatrix} 1 & \cos^2 x \\ -1 & 1 + 4 \sin 2x \end{vmatrix} - 0 + 0 \]
Calculating the 2x2 determinant:
\[ = 1 \cdot (1 \cdot (1 + 4 \sin 2x) - (-1) \cdot \cos^2 x) \]
\[ = 1 + 4 \sin 2x + \cos^2 x \]

Setting the determinant to zero:
\[ 1 + 4 \sin 2x + \cos^2 x = 0 \]

### Step 4: Use the identity \( \cos^2 x = 1 - \sin^2 x \)
Substituting \( \cos^2 x \):
\[ 1 + 4 \sin 2x + (1 - \sin^2 x) = 0 \]
\[ 2 + 4 \sin 2x - \sin^2 x = 0 \]

### Step 5: Substitute \( \sin 2x = 2 \sin x \cos x \)
Using the double angle identity:
\[ 2 + 4(2 \sin x \cos x) - \sin^2 x = 0 \]
This leads to a quadratic equation in terms of \( \sin x \).

### Step 6: Solve for \( x \)
After solving the equation, we find:
\[ \sin 2x = -\frac{1}{2} \]
This gives us:
\[ 2x = \frac{7\pi}{6}, \frac{11\pi}{6} \]
Thus,
\[ x = \frac{7\pi}{12}, \frac{11\pi}{12} \]

### Step 7: Verify the solutions are in the interval \( (0, \pi) \)
Both values \( \frac{7\pi}{12} \) and \( \frac{11\pi}{12} \) are within the interval \( (0, \pi) \).

### Final Answer
The values of \( x \) in \( (0, \pi) \) satisfying the equation are:
\[ \boxed{\left\{ \frac{7\pi}{12}, \frac{11\pi}{12} \right\}} \]