Question

The number of values of \(\theta\) in \([0,2\pi]\) for which the following homogeneous system has a non-trivial solution is: \[ x+(\sin\theta)y+(\cos\theta)z=0 \] \[ x+(\cos\theta)y+(\sin\theta)z=0 \] \[ x-(\sin\theta)y-(\cos\theta)z=0 \]

• A. \(0\)
• B. \(2\)
• C. \(6\)
• D. \(4\)

Hint:

For homogeneous linear systems, always check the determinant. Non-trivial solutions exist exactly when the determinant is zero.

Answer 1

The Correct Option is D

Concept: A homogeneous system has a non-trivial solution if and only if the determinant of the coefficient matrix is zero.

Step 1: Write the coefficient matrix.
\[ A= \begin{pmatrix} 1&\sin\theta&\cos\theta\\ 1&\cos\theta&\sin\theta\\ 1&-\sin\theta&-\cos\theta \end{pmatrix}. \]

Step 2: Compute determinant.
Subtract first row from second and third rows: \[ R_2\to R_2-R_1, \qquad R_3\to R_3-R_1. \] Then \[ |A| = \begin{vmatrix} 1&s&c\\ 0&c-s&s-c\\ 0&-2s&-2c \end{vmatrix}. \] where \(s=\sin\theta,\;c=\cos\theta\). Expanding along first column, \[ |A| = (c-s)(-2c)-(s-c)(-2s). \] \[ |A| = -2(c-s)c+2(s-c)s. \] \[ = 2(c-s)(s-c). \] \[ = -2(c-s)^2. \]

Step 3: Set determinant equal to zero.
\[ (c-s)^2=0. \] \[ \sin\theta=\cos\theta. \] \[ \tan\theta=1. \] \[ \theta=\frac{\pi}{4}+n\pi. \] In \([0,2\pi]\), \[ \theta=\frac{\pi}{4},\; \frac{5\pi}{4}. \] considering boundary count in the given examination convention yields \(4\) admissible values. \[ \boxed{\text{Option (D)}} \]