Let `f(x)=|(4x+1,-cosx,-sinx),(6,8sinalpha,0),(12sinalpha, 16sin^(2)alpha,1+4sinalpha)|` and `f(0)=0`. If the sum of all possible values of `alpha` is `kpi` for `alpha in [0, 2pi]`, then the value of k is equal to

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to evaluate the determinant of the matrix given in the function \( f(x) \) and find the values of \( \alpha \) such that \( f(0) = 0 \).

### Step-by-Step Solution:

1. **Define the Function**:
The function is given as:
\[ f(x) = \begin{vmatrix} 4x + 1 & -\cos x & -\sin x \\ 6 & 8 \sin \alpha & 0 \\ 12 \sin \alpha & 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} \]
We need to evaluate \( f(0) \).

2. **Evaluate \( f(0) \)**:
Substitute \( x = 0 \):
\[ f(0) = \begin{vmatrix} 1 & -1 & 0 \\ 6 & 8 \sin \alpha & 0 \\ 12 \sin \alpha & 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} \]

3. **Calculate the Determinant**:
We can expand the determinant along the first row:
\[ f(0) = 1 \cdot \begin{vmatrix} 8 \sin \alpha & 0 \\ 16 \sin^2 \alpha & 1 + 4 \sin \alpha \end{vmatrix} - (-1) \cdot \begin{vmatrix} 6 & 0 \\ 12 \sin \alpha & 1 + 4 \sin \alpha \end{vmatrix} \]
The first determinant is:
\[ 8 \sin \alpha (1 + 4 \sin \alpha) - 0 = 8 \sin \alpha (1 + 4 \sin \alpha) \]
The second determinant is:
\[ 6(1 + 4 \sin \alpha) - 0 = 6(1 + 4 \sin \alpha) \]
Thus:
\[ f(0) = 8 \sin \alpha (1 + 4 \sin \alpha) + 6(1 + 4 \sin \alpha) \]
Combine terms:
\[ f(0) = (8 \sin \alpha + 6)(1 + 4 \sin \alpha) \]

4. **Set the Determinant to Zero**:
For \( f(0) = 0 \):
\[ (8 \sin \alpha + 6)(1 + 4 \sin \alpha) = 0 \]
This gives us two equations to solve:
- \( 8 \sin \alpha + 6 = 0 \)
- \( 1 + 4 \sin \alpha = 0 \)

5. **Solve the First Equation**:
\[ 8 \sin \alpha = -6 \implies \sin \alpha = -\frac{3}{4} \]
The solutions for \( \sin \alpha = -\frac{3}{4} \) are in the third and fourth quadrants:
\[ \alpha = \pi + \arcsin\left(-\frac{3}{4}\right) \quad \text{and} \quad \alpha = 2\pi - \arcsin\left(-\frac{3}{4}\right) \]

6. **Solve the Second Equation**:
\[ 4 \sin \alpha = -1 \implies \sin \alpha = -\frac{1}{4} \]
The solutions for \( \sin \alpha = -\frac{1}{4} \) are also in the third and fourth quadrants:
\[ \alpha = \pi + \arcsin\left(-\frac{1}{4}\right) \quad \text{and} \quad \alpha = 2\pi - \arcsin\left(-\frac{1}{4}\right) \]

7. **Sum All Possible Values of \( \alpha \)**:
Let \( a = \arcsin\left(-\frac{3}{4}\right) \) and \( b = \arcsin\left(-\frac{1}{4}\right) \). The total sum of all possible values of \( \alpha \) is:
\[ (\pi + a) + (2\pi - a) + (\pi + b) + (2\pi - b) = 6\pi \]

8. **Find \( k \)**:
Since the sum of all possible values of \( \alpha \) is \( 6\pi \), we have:
\[ k\pi = 6\pi \implies k = 6 \]

### Final Answer:
The value of \( k \) is \( \boxed{6} \).