The number of distinct real roots of the equation, | cases x & x &...

Given equation \left|\begin{cases}\cos x & \sin x & \sin x \\ \sin x & \cos x & \sin x \\ \sin x & \sin x & \cos x\end{cases}\right| = 0

Performing row and column operations

${R}_{1} \to {R}_{1} - {R}_{2}$

${R}_{2} \to {R}_{2} - {R}_{3}$

⇒\left|\begin{cases}\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{cases}\right| = 0

${C}_{2} \to {C}_{2} + {C}_{3}$

⇒\left|\begin{cases}\cos x - \sin x & \sin x - \cos x & 0 \\ 0 & 0 & \sin x - \cos x \\ \sin x & \sin x + \cos x & \cos x\end{cases}\right| = 0

Expanding using second row, we get

\left(\sin x + \cos x\right)\left(\cos x - \sin x\right) - \sin x\left(\sin x - \cos x\right) = 0$</p><p>$⇒ \left(\cos x - \sin x\right)\left(\sin x + \cos x + \sin x\right) = 0$</p><p>$⇒ \left(\cos x - \sin x\right)\left(2\sin x + \cos x\right) = 0$</p><p>$⇒ \left(\cos x - \sin x\right) = 0$</p><p>$\sin x = \cos x ⇒ \dfrac{\pi}{4}$</p><p>Or</p><p>$2\sin x + \cos x = 0$</p><p>$\tan x = - \dfrac{1}{2}$</p><p>$x = \tan^{- 1}\left(- \dfrac{1}{2}\right)$</p><p>Hence,$2$ solutions are there.