If `Delta_(1) =|{:(x, sin theta, cos theta),(-sin theta, -x, 1),(cos theta, 1,x):}|` and `Delta_(2) =|{:(x, sin 2theta, cos 2theta),(-sin 2theta, -x, 1),(cos 2theta, 1,x):}|, x ne 0`, then for all `theta in (0, pi/2)`, Then, `Delta_(1) + Delta_(2) =-2x^(k)`. The value of k is ________.

Explanation

To solve the problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and then find the value of \( k \) such that \( \Delta_1 + \Delta_2 = -2x^k \).

### Step 1: Calculate \( \Delta_1 \)

Given:
\[ \Delta_1 = \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we can expand \( \Delta_1 \):
\[ \Delta_1 = x \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} - \sin \theta \begin{vmatrix} -\sin \theta & 1 \\ \cos \theta & x \end{vmatrix} + \cos \theta \begin{vmatrix} -\sin \theta & -x \\ \cos \theta & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} = -x^2 - 1 \)

2. \( \begin{vmatrix} -\sin \theta & 1 \\ \cos \theta & x \end{vmatrix} = -\sin \theta \cdot x - \cos \theta \cdot 1 = -x \sin \theta - \cos \theta \)

3. \( \begin{vmatrix} -\sin \theta & -x \\ \cos \theta & 1 \end{vmatrix} = -\sin \theta \cdot 1 - (-x) \cdot \cos \theta = -\sin \theta + x \cos \theta \)

Substituting back into the determinant:
\[ \Delta_1 = x(-x^2 - 1) + \sin \theta (x \sin \theta + \cos \theta) + \cos \theta (-\sin \theta + x \cos \theta) \]
\[ = -x^3 - x + x \sin^2 \theta + \sin \theta \cos \theta - \cos \theta \sin \theta + x \cos^2 \theta \]
\[ = -x^3 - x + x(\sin^2 \theta + \cos^2 \theta) \]
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \Delta_1 = -x^3 - x + x = -x^3 \]

### Step 2: Calculate \( \Delta_2 \)

Given:
\[ \Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix} \]

Following similar steps as above:
\[ \Delta_2 = x \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} - \sin 2\theta \begin{vmatrix} -\sin 2\theta & 1 \\ \cos 2\theta & x \end{vmatrix} + \cos 2\theta \begin{vmatrix} -\sin 2\theta & -x \\ \cos 2\theta & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} = -x^2 - 1 \)

2. \( \begin{vmatrix} -\sin 2\theta & 1 \\ \cos 2\theta & x \end{vmatrix} = -\sin 2\theta \cdot x - \cos 2\theta \cdot 1 = -x \sin 2\theta - \cos 2\theta \)

3. \( \begin{vmatrix} -\sin 2\theta & -x \\ \cos 2\theta & 1 \end{vmatrix} = -\sin 2\theta \cdot 1 - (-x) \cdot \cos 2\theta = -\sin 2\theta + x \cos 2\theta \)

Substituting back into the determinant:
\[ \Delta_2 = x(-x^2 - 1) + \sin 2\theta (x \sin 2\theta + \cos 2\theta) + \cos 2\theta (-\sin 2\theta + x \cos 2\theta) \]
\[ = -x^3 - x + x \sin^2 2\theta + \sin 2\theta \cos 2\theta - \cos 2\theta \sin 2\theta + x \cos^2 2\theta \]
\[ = -x^3 - x + x(\sin^2 2\theta + \cos^2 2\theta) \]
Using \( \sin^2 2\theta + \cos^2 2\theta = 1 \):
\[ \Delta_2 = -x^3 - x + x = -x^3 \]

### Step 3: Combine \( \Delta_1 \) and \( \Delta_2 \)

Now we add \( \Delta_1 \) and \( \Delta_2 \):
\[ \Delta_1 + \Delta_2 = -x^3 - x^3 = -2x^3 \]

### Step 4: Find the value of \( k \)

From the expression \( \Delta_1 + \Delta_2 = -2x^k \), we have:
\[ -2x^3 = -2x^k \]
Thus, \( k = 3 \).

### Final Answer
The value of \( k \) is \( \boxed{3} \).