The number of distinct real roots of the equation, $\begin{vmatrix} \cos x & \sin x & \sin x \\ \sin x &…
The number of distinct real roots of the equation, $\begin{vmatrix} \cos x & \sin x & \sin x \\ \sin x & \cos x & \sin x \\ \sin x & \sin x & \cos x \end{vmatrix} = 0$ in the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$ is :
Solution
Given equation $\begin{vmatrix} \cos x & \sin x & \sin x \\ \sin x & \cos x & \sin x \\ \sin x & \sin x & \cos x \end{vmatrix} = 0$
Performing row and column operations
$R_1 \rightarrow R_1 - R_2$
$R_2 \rightarrow R_2 - R_3$
$\Rightarrow \begin{vmatrix} \cos x - \sin x & \sin x - \cos x & 0 \\ 0 & \cos x - \sin x & \sin x - \cos x \\ \sin x & \sin x & \cos x \end{vmatrix} = 0$
$C_2 \rightarrow C_2 + C_3$
$\Rightarrow \begin{vmatrix} \cos x - \sin x & \sin x - \cos x & 0 \\ 0 & 0 & \sin x - \cos x \\ \sin x & \sin x + \cos x & \cos x \end{vmatrix} = 0$
Expanding using second row, we get
$|\sin x + \cos x| \cdot |\cos x - \sin x| - \sin x \cdot |\sin x - \cos x| = 0$
$\Rightarrow |\cos x - \sin x| \cdot |\sin x + \cos x + \sin x| = 0$
$\Rightarrow |\cos x - \sin x| \cdot |2\sin x + \cos x| = 0$
$\Rightarrow |\cos x - \sin x| = 0$
$\sin x = \cos x \Rightarrow \frac{\pi}{4}$
Or
$2\sin x + \cos x = 0$
$\tan x = -\frac{1}{2}$
$x = \tan^{-1}\left(-\frac{1}{2}\right)$
Hence, 2 solutions are there.