Question

The sum of all possible values of \( \theta \in [0,2\pi] \), for which the system of equations : \[ x\cos3\theta - 8y - 12z = 0 \] \[ x\cos2\theta + 3y + 3z = 0 \] \[ x + y + 3z = 0 \] has a non-trivial solution, is equal to :

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

Answer 1

The Correct Option is B

Concept:
A homogeneous system of linear equations has a non-trivial solution if the determinant of the coefficient matrix is zero. \[ |A| = 0 \] This condition ensures that the equations are linearly dependent.

Step 1: Form the coefficient matrix.
\[ \begin{vmatrix} \cos3\theta & -8 & -12 \\ \cos2\theta & 3 & 3 \\ 1 & 1 & 3 \end{vmatrix} = 0 \] Step 2: Expand the determinant.
\[ \cos3\theta \begin{vmatrix} 3 & 3 \\ 1 & 3 \end{vmatrix} + 8 \begin{vmatrix} \cos2\theta & 3 \\ 1 & 3 \end{vmatrix} - 12 \begin{vmatrix} \cos2\theta & 3 \\ 1 & 1 \end{vmatrix} = 0 \] Now compute minors:
\[ \begin{vmatrix} 3 & 3 \\ 1 & 3 \end{vmatrix} = 6 \] \[ \begin{vmatrix} \cos2\theta & 3 \\ 1 & 3 \end{vmatrix} = 3\cos2\theta - 3 \] \[ \begin{vmatrix} \cos2\theta & 3 \\ 1 & 1 \end{vmatrix} = \cos2\theta - 3 \] Substitute:
\[ 6\cos3\theta + 8(3\cos2\theta - 3) - 12(\cos2\theta - 3) = 0 \] Step 3: Simplify the equation.
\[ 6\cos3\theta + 24\cos2\theta - 24 - 12\cos2\theta + 36 = 0 \] \[ 6\cos3\theta + 12\cos2\theta + 12 = 0 \] Divide by \(6\):
\[ \cos3\theta + 2\cos2\theta + 2 = 0 \] Step 4: Solve the trigonometric equation.
Using identities:
\[ \cos3\theta = 4\cos^3\theta - 3\cos\theta \] \[ \cos2\theta = 2\cos^2\theta - 1 \] Substituting and simplifying gives:
\[ \cos\theta(\cos\theta + 1)^2 = 0 \] Thus,
\[ \cos\theta = 0 \quad \text{or} \quad \cos\theta = -1 \] Step 5: Find values of \( \theta \) in \( [0,2\pi] \).
\[ \cos\theta = 0 \Rightarrow \theta = \frac{\pi}{2}, \frac{3\pi}{2} \] \[ \cos\theta = -1 \Rightarrow \theta = \pi \] Sum:
\[ \frac{\pi}{2} + \frac{3\pi}{2} + \pi = 2\pi \]