The value of $\theta$ lying between $\theta = 0$ and $\theta = \frac{\pi}{2}$ and satisfying the equation : $$\begin{vmatrix} 1 + \cos^2\theta & \sin^2\theta & 4\sin 4\theta \\ \cos^2\theta & 1 + \sin^2\theta & 4\sin 4\theta \\ \cos^2\theta & \sin^2\theta & 1 + 4\sin 4\theta \end{vmatrix} = 0$$ is :
Step-by-Step Solution
Key Concept: Row operations on determinants preserve structure; solving trigonometric equations requires careful handling of the general solution formula.
Applying operations $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$, the determinant equation yields $(1 + 4\sin 4\theta + \sin^2\theta) - \cos^2\theta = 0$, which simplifies to $1 + 4\sin 4\theta + 1 = 0$, giving $\sin 4\theta = -\frac{1}{2}$. This produces $4\theta = n\pi + (-1)^n(-\frac{\pi}{6})$, so $\theta = \frac{n\pi}{4} + \frac{(-1)^n\pi}{24}$. For $0 \leq \theta \leq \frac{\pi}{2}$, we get $\theta = \frac{7\pi}{24}, \frac{11\pi}{24}$.
Correct Answer: 1,2