Question

If \( \theta_1 \) and \( \theta_2 \) are the values of \( \theta \in (0,\pi) \) for which the system of linear equations \[ x+3y+7z=0, \] \[ -x+4y+7z=0, \] \[ (\sin 3\theta)x+(\cos 2\theta)y+2z=0 \] has a non-trivial solution, then \( |\theta_1-\theta_2| \) is equal to

• A. \( \dfrac{\pi}{6} \)
• B. \( \dfrac{\pi}{3} \)
• C. \( \dfrac{\pi}{2} \)
• D. \( \dfrac{2\pi}{3} \) \bigskip

Hint:

For a homogeneous system of linear equations, always check the determinant of the coefficient matrix. If the determinant is zero, the system admits non-trivial solutions. After obtaining a trigonometric equation, use standard identities such as \[ 1-\cos 2\theta=2\sin^2\theta \] and \[ \sin 3\theta=3\sin\theta-4\sin^3\theta \] to simplify the equation.

Answer 1

The Correct Option is D

Concept: A homogeneous system of linear equations has a non-trivial solution if and only if the determinant of its coefficient matrix is zero. That is, for \[ AX=0, \] a non-trivial solution exists when \[ |A|=0. \]

Step 1: Form the coefficient matrix and equate its determinant to zero. The coefficient matrix is \[ A= \begin{bmatrix} 1 & 3 & 7 -1 & 4 & 7 \sin 3\theta & \cos 2\theta & 2 \end{bmatrix}. \] For a non-trivial solution, \[ \begin{vmatrix} 1 & 3 & 7 -1 & 4 & 7 \sin 3\theta & \cos 2\theta & 2 \end{vmatrix}=0. \] Expanding along the first row, \[ 1 \begin{vmatrix} 4 & 7 \cos 2\theta & 2 \end{vmatrix} -3 \begin{vmatrix} -1 & 7 \sin 3\theta & 2 \end{vmatrix} +7 \begin{vmatrix} -1 & 4 \sin 3\theta & \cos 2\theta \end{vmatrix} =0. \]

Step 2: Evaluate the determinant. \[ (8-7\cos 2\theta) -3(-2-7\sin 3\theta) +7(-\cos 2\theta-4\sin 3\theta) =0. \] \[ 8-7\cos 2\theta +6+21\sin 3\theta -7\cos 2\theta -28\sin 3\theta =0. \] \[ 14-14\cos 2\theta-7\sin 3\theta=0. \] Dividing by \(7\), \[ 2-2\cos 2\theta-\sin 3\theta=0. \] Using \[ 1-\cos 2\theta=2\sin^2\theta, \] we get \[ 4\sin^2\theta-\sin 3\theta=0. \] \[ \sin 3\theta=4\sin^2\theta. \]

Step 3: Solve the trigonometric equation. Using \[ \sin 3\theta = 3\sin\theta-4\sin^3\theta, \] we obtain \[ 3\sin\theta-4\sin^3\theta = 4\sin^2\theta. \] \[ \sin\theta \left( 3-4\sin^2\theta-4\sin\theta \right)=0. \] Since \[ \theta\in(0,\pi), \] \[ \sin\theta\neq 0. \] Therefore, \[ 3-4\sin^2\theta-4\sin\theta=0. \] \[ 4\sin^2\theta+4\sin\theta-3=0. \] Let \[ s=\sin\theta. \] Then \[ 4s^2+4s-3=0. \] \[ (2s-1)(2s+3)=0. \] \[ s=\frac12 \] or \[ s=-\frac32. \] Since \( \sin\theta \in [-1,1] \), \[ \sin\theta=\frac12. \]

Step 4: Find the values of \( \theta \). For \[ \sin\theta=\frac12, \] and \[ \theta\in(0,\pi), \] the solutions are \[ \theta_1=\frac{\pi}{6}, \qquad \theta_2=\frac{5\pi}{6}. \] Hence, \[ |\theta_1-\theta_2| = \left| \frac{\pi}{6} -\frac{5\pi}{6} \right| = \frac{4\pi}{6} = \frac{2\pi}{3}. \]

Step 5: Write the final answer. \[ \boxed{\frac{2\pi}{3}} \]