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: When a determinant contains entries that differ only slightly from one row to another, elementary row operations can greatly simplify the determinant without changing its value. Useful facts: \[ \sin^2\theta+\cos^2\theta=1 \] and \[ \sin 4\theta=\sin(4\theta). \]

Step 1: Apply row operations to simplify the determinant. Let \[ D= \begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin4\theta \sin^2\theta & 1+\cos^2\theta & 4\sin4\theta \sin^2\theta & \cos^2\theta & 1+4\sin4\theta \end{vmatrix}. \] Apply \[ R_1\rightarrow R_1-R_2, \qquad R_3\rightarrow R_3-R_2. \] Then \[ D= \begin{vmatrix} 1 & -1 & 0 \sin^2\theta & 1+\cos^2\theta & 4\sin4\theta 0 & -1 & 1 \end{vmatrix}. \]

Step 2: Evaluate the determinant. Expanding along the first row, \[ D= 1\begin{vmatrix} 1+\cos^2\theta & 4\sin4\theta -1 & 1 \end{vmatrix} - (-1) \begin{vmatrix} \sin^2\theta & 4\sin4\theta 0 & 1 \end{vmatrix}. \] \[ D= \Big((1+\cos^2\theta)+4\sin4\theta\Big) +\sin^2\theta. \] Using \[ \sin^2\theta+\cos^2\theta=1, \] we get \[ D=2+4\sin4\theta. \] Since \(D=0\), \[ 2+4\sin4\theta=0. \] \[ \sin4\theta=-\frac12. \]

Step 3: Solve the trigonometric equation. Since \[ 0<\theta<\frac{\pi}{2}, \] we have \[ 0<4\theta<2\pi. \] Now, \[ \sin4\theta=-\frac12 \] gives \[ 4\theta=\frac{7\pi}{6}, \qquad 4\theta=\frac{11\pi}{6}. \] Therefore, \[ \theta=\frac{7\pi}{24}, \qquad \theta=\frac{11\pi}{24}. \] Both values lie in \[ \left(0,\frac{\pi}{2}\right). \]

Step 4: Write the final answer. \[ \boxed{ \left\{ \frac{7\pi}{24}, \frac{11\pi}{24} \right\} } \]