If `0 lt theta lt pi` and the system of equations
`(sin theta) x + y + z = 0`
`x + (cos theta) y + z = 0`
`(sin theta) x + (cos theta) y + z = 0`
has a non-trivial solution, then `theta =`

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Answer

To solve the given system of equations for non-trivial solutions, we need to find the value of \( \theta \) such that the determinant of the coefficients of the equations equals zero.

The system of equations is:
1. \( (\sin \theta) x + y + z = 0 \)
2. \( x + (\cos \theta) y + z = 0 \)
3. \( (\sin \theta) x + (\cos \theta) y + z = 0 \)

We can represent this system in matrix form \( A \mathbf{x} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \).

The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} \sin \theta & 1 & 1 \\ 1 & \cos \theta & 1 \\ \sin \theta & \cos \theta & 1 \end{bmatrix} \]

To find the values of \( \theta \) for which there are non-trivial solutions, we need to calculate the determinant of matrix \( A \) and set it equal to zero:
\[ \Delta = \det(A) = 0 \]

Calculating the determinant:
\[ \Delta = \begin{vmatrix} \sin \theta & 1 & 1 \\ 1 & \cos \theta & 1 \\ \sin \theta & \cos \theta & 1 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the corresponding elements of the second and third rows.

Substituting the values:
\[ \Delta = \sin \theta \left( \cos \theta \cdot 1 - 1 \cdot \cos \theta \right) - 1 \left( 1 \cdot 1 - 1 \cdot \sin \theta \right) + 1 \left( 1 \cdot \cos \theta - \sin \theta \cdot 1 \right) \]
\[ = \sin \theta (0) - (1 - \sin \theta) + (\cos \theta - \sin \theta) \]
\[ = -1 + \sin \theta + \cos \theta - \sin \theta \]
\[ = \cos \theta - 1 \]

Setting the determinant to zero:
\[ \cos \theta - 1 = 0 \]
\[ \cos \theta = 1 \]

The value of \( \theta \) that satisfies \( \cos \theta = 1 \) is:
\[ \theta = 0 \]
However, since \( 0 < \theta < \pi \), we need to check for other values.

Next, we also consider the case where \( \sin \theta = 1 \):
\[ \sin \theta = 1 \implies \theta = \frac{\pi}{2} \]

Thus, the only value of \( \theta \) in the interval \( (0, \pi) \) that provides a non-trivial solution is:
\[ \theta = \frac{\pi}{2} \]

### Final Answer:
\[ \theta = \frac{\pi}{2} \]