The number of distinct real roots of \(\begin{vmatrix}\cos x & \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x\end{vmatrix} = 0\) in the interval \(\displaystyle -\frac{\pi}{4} \le x \le \frac{\pi}{4}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
Each row has the same sum \(\cos x + 2\sin x\). Applying \(C_1 \to C_1 + C_2 + C_3\) and taking this sum out:
\( \displaystyle \Delta = (\cos x + 2\sin x)\begin{vmatrix}1 & \sin x & \sin x\\ 1 & \cos x & \sin x\\ 1 & \sin x & \cos x\end{vmatrix} \)
Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\( \displaystyle \Delta = (\cos x + 2\sin x)\begin{vmatrix}1 & \sin x & \sin x\\ 0 & \cos x - \sin x & 0\\ 0 & 0 & \cos x - \sin x\end{vmatrix} \)
\(\Delta = (\cos x + 2\sin x)(\cos x - \sin x)^{2}\)
In the interval, \(\cos x \gt 0\), so we may divide by \(\cos x\). The two factors give:
\(\cos x - \sin x = 0\) gives \(\tan x = 1\), so \( \displaystyle x = \frac{\pi}{4} \) (an end point, included)
\(\cos x + 2\sin x = 0\) gives \( \displaystyle \tan x = -\frac{1}{2} \)
On \(\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]\), \(\tan x\) increases from \(-1\) to 1, so it takes the value \(-\frac{1}{2}\) at exactly one point.
So there are 2 distinct roots: \(x = \frac{\pi}{4}\) and the x with \(\tan x = -\frac{1}{2}\) (about \(-0.46\)).
Why the other options are incorrect
Option (1) 0 — Incorrect. \(x = \frac{\pi}{4}\) makes \(\cos x - \sin x = 0\), so there is at least one root.
Option (2) 1 — Incorrect. This misses the root of \(\tan x = -\frac{1}{2}\), which lies inside the interval because \(-1 \lt -\frac{1}{2} \lt 1\).
Option (4) 3 — Incorrect. The squared factor \((\cos x - \sin x)^{2}\) gives the single value \(x = \frac{\pi}{4}\); a repeated factor is not a new root.