The number of distinct real root

Given: $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$

$\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$

$=\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x-\sin x & \sin x-\cos x & \cos x-\cos x \\ \cos x & \cos x & \sin x\end{array}\right|$

$=\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ -(\sin x-\cos x) & \sin x-\cos x & 0 \\ \cos x & \cos x & \sin x\end{array}\right|$

Taking $(\sin x-\cos x)$ common from $R_{2}$

$=(\sin x-\cos x)\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ -1 & 1 & 0 \\ \cos x & \cos x & \sin x\end{array}\right|$

Applying $C_{2} \rightarrow C_{2}+C_{1}$

$=(\sin x-\cos x)\left|\begin{array}{ccc}\sin x & \cos x+\sin x & \cos x \\ -1 & 1-1 & 0 \\ \cos x & \cos x+\cos x & \sin x\end{array}\right|$

$=(\sin x-\cos x)\left|\begin{array}{ccc}\sin x & \cos x+\sin x & \cos x \\ -1 & 0 & 0 \\ \cos x & 2 \cos x & \sin x\end{array}\right|$

Expanding through $R_{2}$

$=(\sin x-\cos x)[-(-1)\{(\cos x+\sin x)(\sin x)-2 \cos x(\cos x)\}]$

$=(\sin x-\cos x)\left[1\left\{\cos x \sin x+\sin ^{2} x-2 \cos ^{2} x\right\}\right]$

$=(\sin x-\cos x)\left(\cos x \sin x+\sin ^{2} x-2 \cos ^{2} x\right)$

$=(\sin x-\cos x)\left(\sin ^{2} x+2 \cos x \sin x-\cos x \sin x-2 \cos ^{2} x\right)$

$=(\sin x-\cos x)(\sin x(\sin x+2 \cos x)-\cos x(\sin x+2 \cos x))$

$=(\sin x-\cos x)(\sin x-\cos x)(\sin x+2 \cos x)$

$=(\sin x-\cos x)^{2}(\sin x+2 \cos x)$

Thus, $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=(\sin x-\cos x)^{2}(\sin x+2 \cos x)$

But it is given that, $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$

$\Rightarrow(\sin x-\cos x)^{2}(\sin x+2 \cos x)=0$

$\Rightarrow(\sin x-\cos x)^{2}=0$ or $\sin x+2 \cos x=0$

$\Rightarrow \sin x-\cos x=0$ or $\sin x=-2 \cos x$

$\Rightarrow \sin x=\cos x$ or $\tan x=-2$

Since $x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$, Thus $\tan x \neq-2$.

$\Rightarrow \sin x=\cos x$ at $x=\frac{\pi}{4}$

Therefore, we have 1 distinct real root.

Hence, the correct option is (c).