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:

\[ \begin{vmatrix} 1 + \sin^2 x & \sin^2 x & \sin^2 x \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4 \sin 2x & 4 \sin 2x & 1 + 4 \sin 2x \end{vmatrix} = 0 \]

we will follow these steps:

### Step 1: Apply Determinant Properties
We will simplify the determinant by applying the property of determinants where we can subtract one column from another. We will perform the following operations:
- Column 1 (C1) = C1 - C3
- Column 2 (C2) = C2 - C3

After applying these operations, the determinant becomes:

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

This simplifies to:

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

### Step 2: Expand the Determinant
Now, we can expand the determinant along the first row:

\[ = 1 \cdot \begin{vmatrix} 1 & \cos^2 x \\ -1 & 1 + 4 \sin 2x \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ = 1 \cdot (1 \cdot (1 + 4 \sin 2x) - (-1) \cdot \cos^2 x) \]
\[ = 1 + 4 \sin 2x + \cos^2 x \]

### Step 3: Set the Determinant to Zero
Now we set the expanded determinant equal to zero:

\[ 1 + 4 \sin 2x + \cos^2 x = 0 \]

Using the identity \(\cos^2 x = 1 - \sin^2 x\):

\[ 1 + 4 \sin 2x + (1 - \sin^2 x) = 0 \]
\[ 2 + 4 \sin 2x - \sin^2 x = 0 \]

### Step 4: Substitute \(\sin 2x\)
Recall that \(\sin 2x = 2 \sin x \cos x\). We can express \(\sin^2 x\) in terms of \(\sin 2x\):

Let \(y = \sin 2x\):

\[ 2 + 4y - \frac{y^2}{4} = 0 \]

Multiplying through by 4 to eliminate the fraction:

\[ 8 + 16y - y^2 = 0 \]
\[ y^2 - 16y - 8 = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = -16\), and \(c = -8\):

\[ y = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]
\[ = \frac{16 \pm \sqrt{256 + 32}}{2} \]
\[ = \frac{16 \pm \sqrt{288}}{2} \]
\[ = \frac{16 \pm 12\sqrt{2}}{2} \]
\[ = 8 \pm 6\sqrt{2} \]

### Step 6: Find Values of \(x\)
Now we have two possible values for \(y\):

1. \(y = 8 + 6\sqrt{2}\) (not valid since \(|y| > 1\))
2. \(y = 8 - 6\sqrt{2}\)

Now, we need to find \(x\) such that:

\[ \sin 2x = 8 - 6\sqrt{2} \]

### Step 7: Determine \(x\) in the Interval \((0, \pi)\)
Since \(\sin 2x\) is periodic, we can find:

\[ 2x = \arcsin(8 - 6\sqrt{2}) + 2k\pi \quad \text{or} \quad 2x = \pi - \arcsin(8 - 6\sqrt{2}) + 2k\pi \]

Dividing by 2 gives us:

\[ x = \frac{1}{2} \arcsin(8 - 6\sqrt{2}) + k\pi \quad \text{or} \quad x = \frac{\pi}{2} - \frac{1}{2} \arcsin(8 - 6\sqrt{2}) + k\pi \]

Considering \(x\) in the interval \((0, \pi)\), we find:

1. \(x = \frac{7\pi}{12}\)
2. \(x = \frac{11\pi}{12}\)

### Final Answer
Thus, the values of \(x\) in the interval \((0, \pi)\) satisfying the equation are:

\[ \boxed{\left\{ \frac{7\pi}{12}, \frac{11\pi}{12} \right\}} \]