Kerala KEAM 5 June 2024 Solved Paper

We are given the determinant of a 3x3 matrix and asked to find the value of $\theta$. First, compute the determinant of the matrix:$$\begin{vmatrix} 0 & -\sin^2 \theta & -2 - 4 \cos 6\theta0 & \cos^2 \theta & -2 - 4 \cos 6\theta1 & \sin \theta & \cos 2\theta \end{vmatrix}.$$We use cofactor expansion along the first row. The determinant simplifies to:$$0 \cdot \begin{vmatrix} \cos^2 \theta & -2 - 4 \cos 6\theta\sin \theta & \cos 2\theta \end{vmatrix} - (-\sin^2 \theta) \cdot \begin{vmatrix} 0 & -2 - 4 \cos 6\theta1 & \cos 2\theta \end{vmatrix} + (-2 - 4 \cos 6\theta) \cdot \begin{vmatrix} 0 & \cos^2 \theta1 & \sin \theta \end{vmatrix}.$$The first term is zero. Now, simplify the second and third terms: $$= \sin^2 \theta \cdot \begin{vmatrix} 0 & -2 - 4 \cos 6\theta1 & \cos 2\theta \end{vmatrix} - (2 + 4 \cos 6\theta) \cdot \begin{vmatrix} 0 & \cos^2 \theta1 & \sin \theta \end{vmatrix}.$$$$\begin{vmatrix} 0 & -2 - 4 \cos 6\theta1 & \cos 2\theta \end{vmatrix} = (0)(\cos 2\theta) - (1)(-2 - 4 \cos 6\theta) = 2 + 4 \cos 6\theta,$$ $$\begin{vmatrix} 0 & \cos^2 \theta1 & \sin \theta \end{vmatrix} = (0)(\sin \theta) - (1)(\cos^2 \theta) = -\cos^2 \theta.$$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 $6\theta$: $$\cos 6\theta = -\frac{1}{2} \quad \Rightarrow \quad 6\theta = \frac{2\pi}{3}, \quad or \quad 6\theta = \frac{4\pi}{3}.$$Step 4: Divide by 6: $$\theta = \frac{\pi}{9}, \quad or \quad \theta = \frac{2\pi}{9}.$$Since $\theta \in \left( 0, \frac{\pi}{3} \right)$, the valid solution is $\theta = \frac{\pi}{9}$.Thus, the correct answer is option (D).Quick Tip: 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.