Consider the system of linear equations in x, y, and z:
`(sin 3 theta) x-y+z=0`
`(cos 2 theta) x+4y+3z=0`
`2x+7y+7z=0`
Which of the following can be the values of `theta` for which the system has a non-trivial solution ?

Options

A. `npi+(-1)^(n) pi//6, AA n in Z`
B. `n pi+(-1)^(n) pi//3, AA n in Z`
C. `npi+(-1)^(n) pi//9, AA n in Z`
D. none of these

Explanation

To solve the given system of linear equations for non-trivial solutions, we need to find the values of \( \theta \) such that the determinant of the coefficient matrix is zero. The equations are:

1. \( \sin(3\theta) x - y + z = 0 \)
2. \( \cos(2\theta) x + 4y + 3z = 0 \)
3. \( 2x + 7y + 7z = 0 \)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) for the system of equations can be represented as:

\[ A = \begin{bmatrix} \sin(3\theta) & -1 & 1 \\ \cos(2\theta) & 4 & 3 \\ 2 & 7 & 7 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Matrix
To find the values of \( \theta \) for which the system has a non-trivial solution, we need to compute the determinant of matrix \( A \) and set it equal to zero:

\[ \text{det}(A) = \begin{vmatrix} \sin(3\theta) & -1 & 1 \\ \cos(2\theta) & 4 & 3 \\ 2 & 7 & 7 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we have:

\[ \text{det}(A) = \sin(3\theta) \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} - (-1) \begin{vmatrix} \cos(2\theta) & 3 \\ 2 & 7 \end{vmatrix} + 1 \begin{vmatrix} \cos(2\theta) & 4 \\ 2 & 7 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} = (4 \cdot 7) - (3 \cdot 7) = 28 - 21 = 7 \)

2. \( \begin{vmatrix} \cos(2\theta) & 3 \\ 2 & 7 \end{vmatrix} = (\cos(2\theta) \cdot 7) - (3 \cdot 2) = 7\cos(2\theta) - 6 \)

3. \( \begin{vmatrix} \cos(2\theta) & 4 \\ 2 & 7 \end{vmatrix} = (\cos(2\theta) \cdot 7) - (4 \cdot 2) = 7\cos(2\theta) - 8 \)

Substituting back into the determinant expression:

\[ \text{det}(A) = \sin(3\theta) \cdot 7 + (7\cos(2\theta) - 6) + (7\cos(2\theta) - 8) \]

Simplifying:

\[ \text{det}(A) = 7\sin(3\theta) + 14\cos(2\theta) - 14 \]

### Step 3: Set the Determinant Equal to Zero
For a non-trivial solution, we set the determinant to zero:

\[ 7\sin(3\theta) + 14\cos(2\theta) - 14 = 0 \]

Dividing the entire equation by 7:

\[ \sin(3\theta) + 2\cos(2\theta) - 2 = 0 \]

### Step 4: Express in Terms of Sine and Cosine
Using the identities \( \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \) and \( \cos(2\theta) = 1 - 2\sin^2(\theta) \):

Substituting these into the equation:

\[ 3\sin(\theta) - 4\sin^3(\theta) + 2(1 - 2\sin^2(\theta)) - 2 = 0 \]

This simplifies to:

\[ 3\sin(\theta) - 4\sin^3(\theta) - 4\sin^2(\theta) = 0 \]

### Step 5: Factor the Equation
Factoring out \( \sin(\theta) \):

\[ \sin(\theta)(-4\sin^2(\theta) + 3 - 4\sin(\theta)) = 0 \]

This gives us:

1. \( \sin(\theta) = 0 \)
2. \( -4\sin^2(\theta) + 3 - 4\sin(\theta) = 0 \)

### Step 6: Solve for \( \theta \)
1. From \( \sin(\theta) = 0 \):
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

2. From \( -4\sin^2(\theta) - 4\sin(\theta) + 3 = 0 \):
Using the quadratic formula, we find:

\[ \sin(\theta) = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot (-4) \cdot 3}}{2 \cdot (-4)} \]

This gives us the solutions for \( \sin(\theta) \).

### Conclusion
The values of \( \theta \) for which the system has a non-trivial solution are:

- \( \theta = n\pi \)
- \( \theta = n\pi \pm \frac{\pi}{6} \) (for \( \sin(\theta) = \frac{1}{2} \))