The number of values of `alpha` in `[-10pi, 10pi]` for which the equations `(sin alpha)x-(cos alpha)y+3z=0, (cos alpha)x+(sin alpha)y-2z=0` and `2x+3y+(cos alpha)z=0` have nontrivial solution is

Options

A. 10
B. 20
C. 40
D. 15

Answer

To determine the number of values of \( \alpha \) in the interval \([-10\pi, 10\pi]\) for which the given equations have nontrivial solutions, we need to analyze the system of equations:

1. \( (\sin \alpha)x - (\cos \alpha)y + 3z = 0 \)
2. \( (\cos \alpha)x + (\sin \alpha)y - 2z = 0 \)
3. \( 2x + 3y + (\cos \alpha)z = 0 \)

### Step 1: Form the Coefficient Matrix
The coefficients of \( x \), \( y \), and \( z \) from the equations can be arranged into a matrix:

\[ A = \begin{bmatrix} \sin \alpha & -\cos \alpha & 3 \\ \cos \alpha & \sin \alpha & -2 \\ 2 & 3 & \cos \alpha \end{bmatrix} \]

### Step 2: Set Up the Determinant
For the system to have nontrivial solutions, the determinant of this matrix must be zero:

\[ \text{det}(A) = 0 \]

### Step 3: Calculate the Determinant
We calculate the determinant using the formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = \sin \alpha \left( \sin \alpha \cdot \cos \alpha - (-2) \cdot 3 \right) - (-\cos \alpha) \left( \cos \alpha \cdot \cos \alpha - (-2) \cdot 2 \right) + 3 \left( \cos \alpha \cdot 3 - 2 \cdot \sin \alpha \right) \]

Calculating each term:

1. \( \sin \alpha (\sin \alpha \cos \alpha + 6) \)
2. \( \cos \alpha (\cos^2 \alpha + 4) \)
3. \( 3(3\cos \alpha - 2\sin \alpha) \)

Putting it all together, we get:

\[ \text{det}(A) = \sin^2 \alpha \cos \alpha + 6 \sin \alpha - \cos^3 \alpha - 4 \cos \alpha + 9 \cos \alpha - 6 \sin \alpha = 0 \]

### Step 4: Simplify the Expression
Combining like terms, we have:

\[ \sin^2 \alpha \cos \alpha - \cos^3 \alpha + 5 \cos \alpha = 0 \]

Factoring out \( \cos \alpha \):

\[ \cos \alpha (\sin^2 \alpha - \cos^2 \alpha + 5) = 0 \]

### Step 5: Solve for \( \alpha \)
This gives us two cases to consider:

1. \( \cos \alpha = 0 \)
2. \( \sin^2 \alpha - \cos^2 \alpha + 5 = 0 \)

#### Case 1: \( \cos \alpha = 0 \)
This occurs at:

\[ \alpha = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]

In the interval \([-10\pi, 10\pi]\), the values of \( n \) can be calculated:

- For \( n = -20 \) to \( n = 20 \), there are \( 20 + 1 = 21 \) values.

#### Case 2: \( \sin^2 \alpha - \cos^2 \alpha + 5 = 0 \)
This simplifies to:

\[ \sin^2 \alpha + 5 = \cos^2 \alpha \]

This equation does not yield any real solutions since \( \sin^2 \alpha + 5 \) is always greater than \( \cos^2 \alpha \) for all real \( \alpha \).

### Final Count of Values
Thus, the only valid solutions come from \( \cos \alpha = 0 \), yielding 21 solutions.

### Conclusion
The total number of values of \( \alpha \) in the interval \([-10\pi, 10\pi]\) for which the equations have nontrivial solutions is:

\[ \boxed{21} \]