Question

The set of all values of \( \theta \) satisfying \(0<\theta<\frac{\pi}{2}\) and \[ \begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin4\theta
\sin^2\theta & 1+\cos^2\theta & 4\sin4\theta
\sin^2\theta & \cos^2\theta & 1+4\sin4\theta \end{vmatrix} =0 \] is

• A. \( \left\{\frac{7\pi}{24}\right\} \)
• B. \( \left\{\frac{11\pi}{24}\right\} \)
• C. \( \left\{\frac{7\pi}{24},\frac{11\pi}{24}\right\} \)
• D. \( \left\{\frac{5\pi}{24},\frac{13\pi}{24}\right\} \) \bigskip

Hint:

For determinants containing many similar rows, first try row operations such as \(R_i-R_j\). This often converts the determinant into a much simpler form and avoids lengthy expansion.

Answer 1

The Correct Option is C

Concept: Use elementary row operations to simplify the determinant before expansion.

Step 1: Apply \(R_1\to R_1-R_2\) and \(R_3\to R_3-R_2\). The determinant reduces to \(2+4\sin4\theta\).

Step 2: Since the determinant is zero, \(2+4\sin4\theta=0\), giving \(\sin4\theta=-\frac12\).

Step 3: As \(0<\theta<\frac{\pi}{2}\), we have \(0<4\theta<2\pi\). Therefore, \(4\theta=\frac{7\pi}{6},\frac{11\pi}{6}\), so \(\boxed{\theta=\frac{7\pi}{24},\frac{11\pi}{24}}\).