The number of distinct real

Solution:

Option (C) 1

Given,

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

Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$, we get

$\left|\begin{array}{lll}2 \cos x+\sin x & \cos x & \cos x \\ 2 \cos x+\sin x & \sin x & \cos x \\ 2 \cos x+\sin x & \cos x & \sin x\end{array}\right|=0$

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

Now,

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

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

$(2 \cos x+\sin x)\left[1 \cdot(\sin x-\cos x)^{2}\right]=0$ (expanding along $C_{1}$ )

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

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

$\tan x=-2$, which is not possible as for $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$,

we get $-1 \leq \tan x \leq 1$

or, $\tan x=1$

Thus, $x=\frac{\pi}{4}$

Therefore, only one real root exist.