We have, $\left|\begin{array}{lll}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0,-\frac{\pi }{4}\le x\le \frac{\pi }{4}$

Apply: $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

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

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

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

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

$\therefore x=\frac{\pi }{4}$

Since, $\frac{-\pi }{4}\le x\le \frac{\pi }{4},$ we get $-1\le \tan x\le 1$

$\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4}$

Hence, number of roots is one.