The values of `theta`, `lambda` for which the following equations `sinthetax-costhetay+(lambda+1)z=0` , `costhetax+sinthetay-lambdaz=0` , `lambdax+(lambda+1)y+costhetaz=0`
have non trivial solution, is

Options

A. `theta=npi`, `lambda in R-{0}`
B. `theta=2npi`, `lambda` is any rational number
C. `theta=(2n+1)pi`, `lambda in R^(+)`, `n in I`
D. `theta=(2n+1)(pi)/(2)`, `lambda in R`, `n in I`

Explanation

To find the values of \( \theta \) and \( \lambda \) for which the given system of equations has a non-trivial solution, we need to analyze the determinant of the coefficient matrix formed by the equations. The equations are:

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

### Step 1: Form the Coefficient Matrix
The coefficients of \( x, y, z \) from the equations form the following matrix:

\[ A = \begin{bmatrix} \sin \theta & -\cos \theta & \lambda + 1 \\ \cos \theta & \sin \theta & -\lambda \\ \lambda & \lambda + 1 & \cos \theta \end{bmatrix} \]

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

\[ \Delta = \begin{vmatrix} \sin \theta & -\cos \theta & \lambda + 1 \\ \cos \theta & \sin \theta & -\lambda \\ \lambda & \lambda + 1 & \cos \theta \end{vmatrix} \]

### Step 3: Expand the Determinant
Using the determinant formula, we can expand it along the first row:

\[ \Delta = \sin \theta \begin{vmatrix} \sin \theta & -\lambda \\ \lambda + 1 & \cos \theta \end{vmatrix} + \cos \theta \begin{vmatrix} \cos \theta & -\lambda \\ \lambda & \cos \theta \end{vmatrix} + (\lambda + 1) \begin{vmatrix} \cos \theta & \sin \theta \\ \lambda & \lambda + 1 \end{vmatrix} \]

Calculating these 2x2 determinants:

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

### Step 4: Set the Determinant to Zero
Combining these results, we set \( \Delta = 0 \):

\[ \sin \theta (\sin \theta \cos \theta + \lambda(\lambda + 1)) + \cos \theta (\cos^2 \theta + \lambda^2) + (\lambda + 1)(\cos \theta(\lambda + 1) - \lambda \sin \theta) = 0 \]

### Step 5: Solve for \( \theta \) and \( \lambda \)
This equation is complex, but we can simplify it. We know that for \( \Delta = 0 \), one condition is \( \cos \theta = 0 \), which gives:

\[ \theta = \frac{(2n + 1)\pi}{2}, \quad n \in \mathbb{Z} \]

For the other condition, we can analyze the quadratic in \( \lambda \):

\[ \lambda^2 + \lambda + 1 = 0 \]

This has no real solutions, indicating that \( \lambda \) can take any real value except for the roots of this quadratic.

### Final Answer
Thus, the values of \( \theta \) and \( \lambda \) for which the system has non-trivial solutions are:

- \( \theta = \frac{(2n + 1)\pi}{2} \) for \( n \in \mathbb{Z} \)
- \( \lambda \in \mathbb{R} \)