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 find the number of values of \(\alpha\) in the interval \([-10\pi, 10\pi]\) for which the equations

\[ (\sin \alpha)x - (\cos \alpha)y + 3z = 0, \]
\[ (\cos \alpha)x + (\sin \alpha)y - 2z = 0, \]
\[ 2x + 3y + (\cos \alpha)z = 0 \]

have a nontrivial solution, we need to analyze the determinant of the coefficient matrix formed by these equations.

### Step 1: Form the Coefficient Matrix
The coefficient matrix \(A\) corresponding to the given equations is:

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

### Step 2: Set the Determinant to Zero
For the system to have a nontrivial solution, we need to set the determinant of the matrix \(A\) to zero:

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

### Step 3: Calculate the Determinant
We can calculate the determinant using the first row:

\[ \text{det}(A) = \sin \alpha \begin{vmatrix} \sin \alpha & -2 \\ 3 & \cos \alpha \end{vmatrix} + \cos \alpha \begin{vmatrix} \cos \alpha & -2 \\ 2 & \cos \alpha \end{vmatrix} + 3 \begin{vmatrix} \cos \alpha & \sin \alpha \\ 2 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} \sin \alpha & -2 \\ 3 & \cos \alpha \end{vmatrix} = \sin \alpha \cos \alpha + 6\)

2. \(\begin{vmatrix} \cos \alpha & -2 \\ 2 & \cos \alpha \end{vmatrix} = \cos^2 \alpha + 4\)

3. \(\begin{vmatrix} \cos \alpha & \sin \alpha \\ 2 & 3 \end{vmatrix} = 3 \cos \alpha - 2 \sin \alpha\)

Putting it all together:

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

### Step 4: Simplify the Determinant
Expanding this expression gives:

\[ \sin^2 \alpha \cos \alpha + 6 \sin \alpha + \cos^3 \alpha + 4 \cos \alpha + 9 \cos \alpha - 6 \sin \alpha \]

Combining like terms:

\[ \sin^2 \alpha \cos \alpha + \cos^3 \alpha + 7 \cos \alpha - 6 \sin \alpha = 0 \]

### Step 5: Factor the Equation
We can factor out \(\cos \alpha\):

\[ \cos \alpha (\sin^2 \alpha + \cos^2 \alpha + 7) - 6 \sin \alpha = 0 \]

Since \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\[ \cos \alpha (1 + 7) - 6 \sin \alpha = 0 \]

This simplifies to:

\[ 8 \cos \alpha - 6 \sin \alpha = 0 \]

### Step 6: Solve for \(\alpha\)
Rearranging gives:

\[ \frac{8}{6} = \frac{\sin \alpha}{\cos \alpha} \Rightarrow \tan \alpha = \frac{4}{3} \]

### Step 7: Find General Solutions
The general solutions for \(\alpha\) are:

\[ \alpha = \tan^{-1}\left(\frac{4}{3}\right) + n\pi \]

### Step 8: Count Solutions in the Interval
To find the number of solutions in the interval \([-10\pi, 10\pi]\):

1. The principal value \(\alpha_0 = \tan^{-1}\left(\frac{4}{3}\right)\).
2. The general solution can be expressed as:

\[ \alpha = \tan^{-1}\left(\frac{4}{3}\right) + n\pi \]

3. The range for \(n\) can be calculated as follows:

\[ -10\pi \leq \tan^{-1}\left(\frac{4}{3}\right) + n\pi \leq 10\pi \]

Calculating the bounds for \(n\):

- For the lower bound:
\[ n \geq \frac{-10\pi - \tan^{-1}\left(\frac{4}{3}\right)}{\pi} \]

- For the upper bound:
\[ n \leq \frac{10\pi - \tan^{-1}\left(\frac{4}{3}\right)}{\pi} \]

### Step 9: Calculate the Number of Integer Values of \(n\)
The total number of integer values of \(n\) can be found by taking the difference of the upper and lower bounds and adding 1.

### Conclusion
After calculating, we find that there are \(20\) values of \(\alpha\) in the interval \([-10\pi, 10\pi]\) for which the equations have a nontrivial solution.