Let $\Delta=\begin{vmatrix}sin x & cos x & cos x\\cos x & sin x & cos x\\cos x & cos x & sin x\end{vmatrix}$
Apply $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=\begin{vmatrix}sin x-cos x& cos x-sin x& 0\\0 & sin x-cos x& cos x-sin x\\cos x& cos x&sin x\end{vmatrix}$
Let us take (sin x-cos x) as a common factor from $R_1$ and $R_2$
$\Delta=(sin x-cos x)\begin{vmatrix}1 & -1 & 0\\0 & 1 & -1\\cos x & cos x & sin x\end{vmatrix}$
Now expanding along $R_1$ we get,
$\Delta=(sin x-cos x)^2[1(sin x+cos x)-1(0+cos x)]$
$\quad=(sin x-cos x)^2[sin x+cos x-cos x]$
But it is given $\Delta=0$
($sin x-cos x)^2[sin x]=0.$
If $(sin x-cos x)^2=0.$
$\Rightarrow sin x=cos x=\frac{1}{\sqrt 2}$
This is possible if $x=\frac{\pi}{4}$
and sin x=0.if x=0.
We know $\frac{1}{\sqrt 2}$ is irrational.
Hence sin x=0 is the only distinct real root,between the interval $\frac{-\pi}{4}\leq x\leq \frac{\pi}{4}$
Hence C is the correct answer.