Explanation

(c)
Applying $R_{1} \rightarrow R_{1} + R_{2},\$we get
$\left| \begin{matrix}
2 & 2 & 1 \\
\cos^{2}\theta & 1 + \cos^{2}\theta & \cos^{2}\theta \\
4\sin{4\ \theta} & 4\sin{4\ \theta} & 1 + 4\sin{4\ \theta}
\end{matrix} \right| = 0$
Applying $C_{1} \rightarrow C_{1} - 2C_{3},\ C_{2} \rightarrow C_{2} - 2C_{3}$
$\left| \begin{matrix}
0 & 0 & 1 \\
- \cos^{2}\theta & 1 - \cos^{2}\theta & \cos^{2}\theta \\
- 2 - 4\sin{4\ \theta} & - 2 - 4\sin{4\ \theta} & 1 + 4\sin{4\ \theta}
\end{matrix} \right|$=0
$\Rightarrow \lbrack\cos^{2}{\theta(2 + 4\sin{4\ \theta) + (1 - \cos^{2}{\theta)(2 + 4\sin{4\ \theta)\rbrack = 0}}}}$
$\Rightarrow \lbrack 2\cos^{2}{\theta + 4\cos^{2}{\theta\sin{4\ \theta} + 2 + 4\sin{4\theta}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 2\cos^{2}{\theta - 4\cos^{2}{\theta\sin{4\theta\rbrack = 0}}}$
$\Rightarrow 2 + 4\sin{4\theta = 0}$
$\Rightarrow \ \sin{4\theta = - \frac{1}{2}}$