Explanation

(c)
We have, $\left| \begin{matrix}
\sin x & \cos x & \cos x \\
\cos x & \sin x & \cos x \\
\cos x & \cos x & \sin x
\end{matrix} \right| = 0$
Applying $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$
$\Rightarrow \ \left| \begin{matrix}
\sin x + 2\cos x & \cos x & \cos x \\
\sin x + 2\cos x & \sin x & \cos x \\
\sin x + 2\cos x & \cos x & \sin x
\end{matrix} \right| = 0$
$\Rightarrow \left( 2\cos x + \sin x \right)\left| \begin{matrix}
1 & \cos x & \cos x \\
1 & \sin x & \cos x \\
1 & \cos x & \sin x
\end{matrix} \right| = 0$
Applying $R_{2} \rightarrow R_{2} - R_{1},\ R_{3} \rightarrow R_{3} - R_{1}$
$\Rightarrow \ \left( 2\cos x + \sin x \right)\left| \begin{matrix}
1 & \cos x & \cos x \\
0 & \sin x–\cos x & 0 \\
0 & 0 & \sin x–\cos x
\end{matrix} \right| = 0\$
$\Rightarrow \ \left( 2\cos x + \sin x \right)\left( \sin x–\cos x \right)^{2} = 0$
$\therefore\tan{x = - 2,1\ But\tan{x \neq - 2,}}\$because it does not lie in the interval $\left\lbrack \ - \frac{\pi}{4,}\frac{\pi}{4} \right\rbrack$.
$\therefore\tan{x = 1}$
So, $x = \frac{\pi}{4}$