The number of distinct real roots of the equation, $\left|\begin{array}{ccc}\cos x& \sin x& \sin x\\ \sin x& \cos x& \sin x\\ \sin x& \sin x& \cos x\end{array}\right|=0$ in the interval $\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$ is :

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

Performing row and column operations

$R_1\to R_1-R_2$

$R_2\to R_2-R_3$

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

$C_2\to C_2+C_3$

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

Expanding using second row, we get 

$\left(\sin x+\cos x\right)\left(\cos x-\sin x\right)-\sin x\left(\sin x-\cos x\right)=0$

$\Rightarrow \left(\cos x-\sin x\right)\left(\sin x+\cos x+\sin x\right)=0$

$\Rightarrow \left(\cos x-\sin x\right)\left(2\sin x+\cos x\right)=0$

$\Rightarrow \left(\cos x-\sin x\right)=0$

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

Or

$2\sin x+\cos x=0$

$\tan x=-\frac{1}{2}$

$x={\tan }^{-1}\left(-\frac{1}{2}\right)$

Hence, $2$ solutions are there.