The system of equations
`x - y cos theta + z cos 2theta =0`
`-xcostheta+y-zcostheta=0`
`x cos 2theta-ycostheta + z=0`
has non-trivial solution for `theta` equals

Options

A. `pi/3`
B. `pi/6`
C. `(2pi)/3`
D. `pi/12`

Answer

To solve the given system of equations for non-trivial solutions, we need to analyze the determinant of the coefficients of the variables \(x\), \(y\), and \(z\). The equations are:

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

### Step 1: Write the system in matrix form
We can represent the system of equations in matrix form as follows:

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

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

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

### Step 3: Expand the determinant
Using the determinant formula for a 3x3 matrix, we can expand \(D\):

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

Calculating each of these 2x2 determinants:

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

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

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

Substituting these back into the determinant:

\[ D = \sin^2 \theta + \cos \theta (-\cos \theta + \cos \theta \cos 2\theta) + \cos 2\theta (\cos^2 \theta - \cos 2\theta) \]

### Step 4: Simplify the determinant
Now we simplify \(D\):

\[ D = \sin^2 \theta - \cos^2 \theta + \cos^2 \theta \cos 2\theta + \cos 2\theta \cos^2 \theta - \cos^2 \theta \cos 2\theta \]

This simplifies to:

\[ D = \sin^2 \theta - \cos^2 \theta + \cos^2 \theta \cos 2\theta \]

Using the identity \(\cos 2\theta = 2\cos^2 \theta - 1\):

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

### Step 5: Set the determinant to zero
Now we set \(D = 0\):

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

This can be solved for \(\theta\) to find the values where the determinant is zero, indicating non-trivial solutions.

### Step 6: Solve for \(\theta\)
After simplification, we find that the equation holds for any \(\theta\) such that:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

This is true for all \(\theta\). Thus, the system has a non-trivial solution for all real numbers \(\theta\).

### Final Answer
The system of equations has non-trivial solutions for \(\theta\) belonging to all real numbers.

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