the set of equations `lambdax-y+(cos theta) z=0,3x+y+2z=0`
`(cos theta) x+y+2z=0` ,`0 le theta lt 2 pi` has non-trivial solution (s)

Options

A. A. for no value of `lambda` and `theta`
B. B. for all value of `lambda`and `theta`
C. C. for all value of `lambda` and only two values of `theta`
D. D. for only one value of `lambda` and all values of `theta`

Answer

To determine the conditions under which the given set of equations has non-trivial solutions, we will analyze the determinant of the coefficients of the variables \(x\), \(y\), and \(z\). The equations provided are:

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

### Step 1: Set up the coefficient matrix
We can express the system of equations in matrix form as follows:

\[ \begin{bmatrix} \lambda & -1 & \cos \theta \\ 3 & 1 & 2 \\ \cos \theta & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Calculate the determinant
For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero:

\[ D = \begin{vmatrix} \lambda & -1 & \cos \theta \\ 3 & 1 & 2 \\ \cos \theta & 1 & 2 \end{vmatrix} \]

Calculating the determinant using cofactor expansion:

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

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 0\)

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

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

Putting it all together:

\[ D = \lambda \cdot 0 + (6 - 2\cos \theta) + \cos \theta(3 - \cos \theta) \]

This simplifies to:

\[ D = 6 - 2\cos \theta + 3\cos \theta - \cos^2 \theta \]
\[ D = 6 + \cos \theta - \cos^2 \theta \]

### Step 3: Set the determinant equal to zero
For non-trivial solutions, we set the determinant to zero:

\[ 6 + \cos \theta - \cos^2 \theta = 0 \]

Rearranging gives:

\[ \cos^2 \theta - \cos \theta - 6 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = -1\), and \(c = -6\):

\[ \cos \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]
\[ \cos \theta = \frac{1 \pm \sqrt{1 + 24}}{2} \]
\[ \cos \theta = \frac{1 \pm 5}{2} \]

Calculating the two potential solutions:

1. \(\cos \theta = \frac{6}{2} = 3\) (not possible since \(\cos \theta\) must be between -1 and 1)
2. \(\cos \theta = \frac{-4}{2} = -2\) (also not possible)

### Conclusion
Since both potential values for \(\cos \theta\) are outside the range of the cosine function, there are no values of \(\lambda\) and \(\theta\) that satisfy the condition for non-trivial solutions.

Thus, the final answer is that there are no values of \(\lambda\) and \(\theta\) for which the system has non-trivial solutions.