The number of distinct real roots of $\left|\begin{array}{l}\sin x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos x\\ \cos x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos x\\ \cos x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cos x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sin x\end{array}\right|=0$ in the interval $-\frac{\pi }{4}\le x\le \frac{\pi }{4}$ is 

To simplify the determinant, let $\sin x=a;\cos =b.$ Then the equation becomes $\left|\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\end{array}\right|=0$

$\text{ or }(a-b{)}^2(a-2b)=0\text{ or }a=b\text{ or }a=2b\text{ or }\frac{a}{b}=1\text{ or }$ $\frac{a}{b}=2\Rightarrow \tan x=1\text{ or }\tan x=2$

$\text{ But we have }-\frac{\pi }{4}\le x\le \frac{\pi }{4}\Rightarrow \tan \left(\frac{\pi }{4}\right)\le \tan x\le \left(\frac{\pi }{4}\right)$

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

Therefore , there is only one real root