Question

If system of equations \(x \cos \theta - 8y - 12z = 0\), \(x \cos 2\theta + y + 3z = 0\), \(x + y + 3z = 0\) has non-trivial solution, then find sum of values of \(\theta\) (where \(\theta \in [0, 2\pi]\)).

• A. \(\pi\)
• B. \(2\pi\)
• C. \(3\pi\)
• D. \(4\pi\)

Hint:

Always look for row or column operations to create zeros in a matrix before expanding the determinant. It significantly reduces the complexity of the trigonometric equation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a non-trivial solution in a homogeneous system, the determinant of the coefficient matrix must be zero.

Step 2: Key Formula or Approach:
\[ \begin{vmatrix} \cos \theta & -8 & -12
\cos 2\theta & 1 & 3
1 & 1 & 3 \end{vmatrix} = 0 \]

Step 3: Detailed Explanation:
1. Simplify the determinant by performing \( R_2 \to R_2 - R_3 \): \[ \begin{vmatrix} \cos \theta & -8 & -12
\cos 2\theta - 1 & 0 & 0
1 & 1 & 3 \end{vmatrix} = 0 \] 2. Expand along the second row: \[ -(\cos 2\theta - 1) [(-8)(3) - (1)(-12)] = 0 \] \[ -(\cos 2\theta - 1) [-24 + 12] = 0 \implies 12(\cos 2\theta - 1) = 0 \] 3. Solve for \( \theta \): \[ \cos 2\theta = 1 \] In the interval \( [0, 2\pi] \), \( 2\theta \) can be \( 0, 2\pi, 4\pi \). \[ \theta = 0, \pi, 2\pi \] 4. Sum of values: \( 0 + \pi + 2\pi = 3\pi \). *(Note: If the system implies \( \cos 3\theta \) as per similar problems, the result is usually \( 2\pi \). For the specific equations provided here, the sum is \( 3\pi \)).*

Step 4: Final Answer:
The sum of values of \( \theta \) is \( 3\pi \).