Explanation

(c)
Since, $\left| \begin{matrix}
\sin x & \cos x & \cos x \\
\cos x & \sin x & \cos x \\
\cos x & \cos x & \sin x
\end{matrix} \right| = 0$
$\Rightarrow \sin{x(\sin^{2}{x - \cos^{2}{x) - \cos{x(\cos{x\sin{x - \cos^{2}{x)}}}}}}}$
$+ \cos{x(\cos^{2}{x - \sin{x\cos x) = 0}}}$
$\Rightarrow \sin{x(\sin^{2}{x - \cos^{2}{x) - 2\cos^{2}{x(\sin{x - \cos{x) = 0}}}}}}$
$\Rightarrow (\sin{x - \cos{x)\lbrack\sin{x(\sin{x + \cos{x) - 2\cos^{2}{x\rbrack = 0}}}}}}$
$\Rightarrow (\sin{x - \cos{x)\lbrack(\sin^{2}{x - \cos^{2}{x)}}}}$
$+ (\sin{x\cos x - \cos^{2}{x)\rbrack = 0}}$
$\Rightarrow$ $({\sin{x - \cos{x)}}}^{2}\lbrack\sin{x + \cos{x + \cos{x\rbrack =}}}$0
$\Rightarrow ({\sin{x - \cos{x)}}}^{2}(\sin{x + 2\cos{x) = 0}}$
$\Rightarrow$ Either$\ ({\sin{x - \cos{x)}}}^{2} = 0$
or $\sin{x + 2\cos{x) = 0}}$
$\Rightarrow$Either $\tan{x = 1\ or\tan{x = - 2}}$
$\Rightarrow \$Either $x = \frac{\pi}{4}\ or\ \tan{x = - 2}$
As$\ x \in \left\lbrack - \frac{\pi}{4},\frac{\pi}{4} \right\rbrack$, $\tan{x \in \lbrack - 1,\ 1\rbrack}$
Hence, real solution is only $x = \frac{\pi}{4}$