Question

If \[ \theta \in \left( 0, \frac{\pi}{3} \right) \quad \text{and} \quad \begin{vmatrix} 0 & -\sin^2 \theta & -2 - 4 \cos 6\theta \\ 0 & \cos^2 \theta & -2 - 4 \cos 6\theta \\ 1 & \sin \theta & \cos 2\theta \end{vmatrix} = 0, \] then \( \theta \) is equal to:

• A. \( \frac{\pi}{18} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{\pi}{9} \)
• E. \( \frac{\pi}{5} \)

Hint:

When solving trigonometric equations, use the periodicity of trigonometric functions and their known values to find solutions in a specific interval. For example, the cosine function has known values for specific angles.

Answer 1

The Correct Option is D

We are given the determinant of a 3x3 matrix and asked to find the value of \( \theta \). First, compute the determinant of the matrix:

\[ \begin{vmatrix} 0 & -\sin^2 \theta & -2 - 4 \cos 6\theta \\ 0 & \cos^2 \theta & -2 - 4 \cos 6\theta \\ 1 & \sin \theta & \cos 2\theta \end{vmatrix}. \]
We use cofactor expansion along the first row. The determinant simplifies to:

\[ 0 \cdot \begin{vmatrix} \cos^2 \theta & -2 - 4 \cos 6\theta \\ \sin \theta & \cos 2\theta \end{vmatrix} - (-\sin^2 \theta) \cdot \begin{vmatrix} 0 & -2 - 4 \cos 6\theta \\ 1 & \cos 2\theta \end{vmatrix} + (-2 - 4 \cos 6\theta) \cdot \begin{vmatrix} 0 & \cos^2 \theta \\ 1 & \sin \theta \end{vmatrix}. \]
The first term is zero. Now, simplify the second and third terms:

\[ = \sin^2 \theta \cdot \begin{vmatrix} 0 & -2 - 4 \cos 6\theta \\ 1 & \cos 2\theta \end{vmatrix} - (2 + 4 \cos 6\theta) \cdot \begin{vmatrix} 0 & \cos^2 \theta \\ 1 & \sin \theta \end{vmatrix}. \]
\[ \begin{vmatrix} 0 & -2 - 4 \cos 6\theta \\ 1 & \cos 2\theta \end{vmatrix} = (0)(\cos 2\theta) - (1)(-2 - 4 \cos 6\theta) = 2 + 4 \cos 6\theta, \] \[ \begin{vmatrix} 0 & \cos^2 \theta \\ 1 & \sin \theta \end{vmatrix} = (0)(\sin \theta) - (1)(\cos^2 \theta) = -\cos^2 \theta. \]
Thus, the determinant simplifies to: \[ \sin^2 \theta (2 + 4 \cos 6\theta) - (2 + 4 \cos 6\theta) (-\cos^2 \theta). \] Factor out \( (2 + 4 \cos 6\theta) \): \[ (2 + 4 \cos 6\theta)(\sin^2 \theta + \cos^2 \theta) = 0. \] Since \( \sin^2 \theta + \cos^2 \theta = 1 \), the equation becomes: \[ 2 + 4 \cos 6\theta = 0. \] Step 2: Solve for \( \cos 6\theta \): \[ 4 \cos 6\theta = -2 \quad \Rightarrow \quad \cos 6\theta = -\frac{1}{2}. \] Step 3: Solve for \( 6\theta \): \[ \cos 6\theta = -\frac{1}{2} \quad \Rightarrow \quad 6\theta = \frac{2\pi}{3}, \quad \text{or} \quad 6\theta = \frac{4\pi}{3}. \] Step 4: Divide by 6: \[ \theta = \frac{\pi}{9}, \quad \text{or} \quad \theta = \frac{2\pi}{9}. \] Since \( \theta \in \left( 0, \frac{\pi}{3} \right) \), the valid solution is \( \theta = \frac{\pi}{9} \).

Thus, the correct answer is option (D).