The number of distinct real roots of $\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

$1$

The correct answer is Option (3) → $1$ ##

We have,

$\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$

On expanding along $R_1$, we get

$\sin x [\sin^2 x - \cos^2 x] - \cos x [\sin x \cos x - \cos^2 x] + \cos x [\cos^2 x - \cos x \sin x] = 0$

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

$\Rightarrow (\sin x - \cos x) [\sin^2 x + \sin x \cos x - 2 \cos^2 x] = 0$

$\Rightarrow (\sin x - \cos x) [\sin^2 x + 2 \sin x \cos x - \sin x \cos x - 2 \cos^2 x] = 0$

$\Rightarrow (\sin x - \cos x) [\sin x (\sin x + 2 \cos x) - \cos x (\sin x + 2 \cos x)] = 0$

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

Either  $2 \cos x = -\sin x$

$\Rightarrow \cos x = -\frac{1}{2} \sin x$

$\Rightarrow \tan x = -2$

But here for $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$, we get $-1 \leq \tan x \leq 1$, so no solution possible for $\tan x = -2$.

or $(\sin x - \cos x)^2 = 0$

$\Rightarrow \sin x - \cos x = 0$

$\Rightarrow \sin x = \cos x$

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

$∴x = \frac{\pi}{4}$

So, only one distinct real root exists.