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 has a non-trivial solution only when the determinant of its coefficient matrix is zero.

Step 1: Form the determinant and simplify to obtain \(2-2\cos2\theta-\sin3\theta=0\). Using \(1-\cos2\theta=2\sin^2\theta\), this becomes \(4\sin^2\theta=\sin3\theta\).

Step 2: Substituting \(\sin3\theta=3\sin\theta-4\sin^3\theta\) gives \(4\sin^2\theta+4\sin\theta-3=0\). Let \(s=\sin\theta\). Then \((2s-1)(2s+3)=0\), so \(s=\frac12\).

Step 3: Hence, \(\theta=\frac{\pi}{6},\frac{5\pi}{6}\), giving \(\left|\theta_1-\theta_2\right|=\boxed{\frac{2\pi}{3}}\).