Question

Let \(\theta \in \left(0, \frac{\pi}{2}\right)\). If the system of linear equations
\((1 + \cos^2\theta)x + \sin^2\theta y + 4\sin3\theta z = 0\)
\(\cos^2\theta x + (1 + \sin^2\theta)y + 4\sin3\theta z = 0\)
\(\cos^2\theta x + \sin^2\theta y + (1 + 4\sin3\theta)z = 0\)
has a non-trivial solution, then the value of \(\theta\) is :

• A. \(\frac{\pi}{18}\)
• B. \(\frac{7\pi}{18}\)
• C. \(\frac{5\pi}{18}\)
• D. \(\frac{4\pi}{9}\)

Hint:

In systems involving trigonometric identities like \(\sin^2\theta\) and \(\cos^2\theta\), subtracting adjacent rows often simplifies the matrix significantly to reveal the identity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A homogeneous system of linear equations has a non-trivial solution if and only if the determinant of its coefficient matrix is zero.
Step 2: Key Formula or Approach:
Set \(\Delta = 0\), where \(\Delta\) is the determinant of the coefficients. Use row and column operations to simplify the determinant.
Step 3: Detailed Explanation:
The determinant is:

\[ \Delta = \begin{vmatrix} 1 + \cos^2\theta & \sin^2\theta & 4\sin3\theta \\ \cos^2\theta & 1 + \sin^2\theta & 4\sin3\theta \\ \cos^2\theta & \sin^2\theta & 1 + 4\sin3\theta \end{vmatrix} = 0 \]

Perform \(R_1 \to R_1 - R_2\) and \(R_2 \to R_2 - R_3\):

\[ \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \cos^2\theta & \sin^2\theta & 1 + 4\sin3\theta \end{vmatrix} = 0 \]

Expanding along \(R_1\):

\[ 1(1 + 4\sin3\theta + \sin^2\theta) - (-1)(0 + \cos^2\theta) = 0 \]

\[ 1 + 4\sin3\theta + \sin^2\theta + \cos^2\theta = 0 \]

Using \(\sin^2\theta + \cos^2\theta = 1\):

\[ 1 + 4\sin3\theta + 1 = 0 \implies 4\sin3\theta = -2 \implies \sin3\theta = -\frac{1}{2} \]

Since \(\theta \in (0, \pi/2)\), we have \(3\theta \in (0, 3\pi/2)\).
The value \(\sin x = -1/2\) in this range occurs at \(x = \pi + \pi/6 = \frac{7\pi}{6}\).

\[ 3\theta = \frac{7\pi}{6} \implies \theta = \frac{7\pi}{18} \]

Step 4: Final Answer:
The value of \(\theta\) is \(\frac{7\pi}{18}\).