Explanation

(a)
Applying $C_{1} \rightarrow C_{1} + C_{2},$we get
$\Rightarrow \left| \begin{matrix}
2 & \cos^{2}\theta & 4\sin{4\ \theta} \\
2 & 1 + \cos^{2}\theta & 4\sin{4\ \theta} \\
1 & \cos^{2}\theta & 1 + 4\sin{4\ \theta}
\end{matrix} \right| = 0$
$\Rightarrow \left| \begin{matrix}
2 & \cos^{2}\theta & 4\sin{4\ \theta} \\
0 & 1 & 0 \\
1 & \cos^{2}\theta & 1 + 4\sin{4\ \theta}
\end{matrix} \right| = 0$
[$R_{2} \rightarrow R_{2} - R_{1}\rbrack$
$\Rightarrow \ \ \ (2 + 4\sin{4\ \theta) = 0}$
$\Rightarrow \ \ \sin{4\theta = - \frac{1}{2} = - \sin\frac{\pi}{6}}$
$\Rightarrow \ \ \ \ 4\ \theta = n\pi + ( - 1)^{n}\left( - \frac{\pi}{6} \right)$
$\therefore$ The value of $\theta$ between 0 and $\frac{\pi}{2}$ will be$\ \frac{7\pi}{24}$ and $\frac{11\pi}{24}$