If `0lt=thetalt=pi` and the system of equations
`x=(sin theta)y+(cos theta)z`
`y=z+(cos theta) x`
`z=(sin theta) x+y`
has a non-trivial solution then `(8theta)/(pi)` is equal to

Answer

To solve the given system of equations for a non-trivial solution, we will follow these steps:

### Step 1: Write the system of equations
The given equations are:
1. \( x = \sin(\theta)y + \cos(\theta)z \)
2. \( y = z + \cos(\theta)x \)
3. \( z = \sin(\theta)x + y \)

### Step 2: Rearrange the equations
We can rearrange these equations to express them in a standard form:
1. \( -\sin(\theta)y - \cos(\theta)z + x = 0 \)
2. \( -\cos(\theta)x + y - z = 0 \)
3. \( -\sin(\theta)x - y + z = 0 \)

### Step 3: Form the coefficient matrix
The coefficient matrix \( A \) for the system can be written as:
\[ A = \begin{pmatrix} 1 & -\sin(\theta) & -\cos(\theta) \\ -\cos(\theta) & 1 & -1 \\ -\sin(\theta) & -1 & 1 \end{pmatrix} \]

### Step 4: Find the determinant of the matrix
To find a non-trivial solution, we need to set the determinant of the matrix \( A \) to zero:
\[ \text{det}(A) = 0 \]

Calculating the determinant:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} - (-\sin(\theta)) \cdot \begin{vmatrix} -\cos(\theta) & -1 \\ -\sin(\theta) & 1 \end{vmatrix} - \cos(\theta) \cdot \begin{vmatrix} -\cos(\theta) & 1 \\ -\sin(\theta) & -1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} = 1 \cdot 1 - (-1)(-1) = 1 - 1 = 0 \)

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

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

Substituting back into the determinant:
\[ \text{det}(A) = 1 \cdot 0 + \sin(\theta)(\cos(\theta) + \sin(\theta)) - \cos(\theta)(\cos(\theta) + \sin(\theta)) \]
\[ = \sin(\theta)(\cos(\theta) + \sin(\theta)) - \cos(\theta)(\cos(\theta) + \sin(\theta)) \]
\[ = (\sin(\theta) - \cos(\theta))(\cos(\theta) + \sin(\theta)) \]

### Step 5: Set the determinant to zero
For a non-trivial solution, we set:
\[ (\sin(\theta) - \cos(\theta))(\cos(\theta) + \sin(\theta)) = 0 \]

This gives us two cases:
1. \( \sin(\theta) - \cos(\theta) = 0 \) → \( \tan(\theta) = 1 \) → \( \theta = \frac{\pi}{4} \)
2. \( \cos(\theta) + \sin(\theta) = 0 \) → \( \sin(\theta) = -\cos(\theta) \) → \( \theta = \frac{3\pi}{4} \)

### Step 6: Evaluate \( \frac{8\theta}{\pi} \)
We will take \( \theta = \frac{3\pi}{4} \) as it falls within the range \( 0 < \theta < \pi \):
\[ \frac{8\theta}{\pi} = \frac{8 \cdot \frac{3\pi}{4}}{\pi} = \frac{24\pi}{4\pi} = 6 \]

### Final Answer
Thus, the value of \( \frac{8\theta}{\pi} \) is \( 6 \).

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