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 determine the values of \(\theta\) for which the given system of linear equations has a non-trivial solution, we need to analyze the determinant of the coefficients of the equations. 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: Write the Coefficient Matrix
We can express the system in matrix form as follows:

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

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

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

### Step 3: Expand the Determinant
Using the determinant formula, we can expand it as follows:

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

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 these back into the determinant:

\[ D = \sin(3\theta) \cdot 7 + (7\cos(2\theta) - 6) + (7\cos(2\theta) - 8) \]

Combining the terms:

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

### Step 4: Set the Determinant to Zero
For the system to have a non-trivial solution, we set the determinant to zero:

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

Dividing through by 7:

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

### Step 5: Express \(\sin(3\theta)\) and \(\cos(2\theta)\)
Using the identities:

- \(\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)\)
- \(\cos(2\theta) = 1 - 2\sin^2(\theta)\)

Substituting these into the equation gives:

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

Simplifying, we have:

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

### Step 6: Factor the Equation
Factoring out \(\sin(\theta)\):

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

This gives us:

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

### Step 7: Solve the Quadratic Equation
Using the quadratic formula for the second equation:

\[ \sin(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 48}}{-8} = \frac{-4 \pm 8}{-8} \]

Calculating the roots:

1. \(\sin(\theta) = \frac{1}{2}\)
2. \(\sin(\theta) = -\frac{3}{2}\) (not possible)

### Step 8: Find \(\theta\)
From \(\sin(\theta) = 0\):

\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

From \(\sin(\theta) = \frac{1}{2}\):

\[ \theta = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2k\pi \quad (k \in \mathbb{Z}) \]

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

\[ \theta = n\pi \quad \text{and} \quad \theta = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2k\pi \]