Question:
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
\[
\left(-\frac{\pi}{4},\frac{\pi}{4}\right)
\]
is
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
\[
\left(-\frac{\pi}{4},\frac{\pi}{4}\right)
\]
is
Show Hint
For matrices with equal diagonal entries and equal off-diagonal entries, use the standard determinant formula directly.
Updated On: Jun 3, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Concept
For a matrix having diagonal entries $a$ and off-diagonal entries $b$, \[ \det=(a-b)^2(a+2b). \]
Step 2: Meaning
Here \[ a=\sin x,\qquad b=\cos x. \] Therefore \[ \det=(\sin x-\cos x)^2(\sin x+2\cos x). \]
Step 3: Analysis
Setting determinant equal to zero, \[ (\sin x-\cos x)^2(\sin x+2\cos x)=0. \] Thus \[ \sin x=\cos x \] or \[ \sin x+2\cos x=0. \] The first gives \[ \tan x=1 \Rightarrow x=\frac{\pi}{4}, \] which is not included in the interval. The second gives \[ \tan x=-2. \] This has exactly one solution in \[ \left(-\frac{\pi}{4},\frac{\pi}{4}\right). \]
Step 4: Conclusion
Hence there is exactly one real root.
Final Answer: (B)
For a matrix having diagonal entries $a$ and off-diagonal entries $b$, \[ \det=(a-b)^2(a+2b). \]
Step 2: Meaning
Here \[ a=\sin x,\qquad b=\cos x. \] Therefore \[ \det=(\sin x-\cos x)^2(\sin x+2\cos x). \]
Step 3: Analysis
Setting determinant equal to zero, \[ (\sin x-\cos x)^2(\sin x+2\cos x)=0. \] Thus \[ \sin x=\cos x \] or \[ \sin x+2\cos x=0. \] The first gives \[ \tan x=1 \Rightarrow x=\frac{\pi}{4}, \] which is not included in the interval. The second gives \[ \tan x=-2. \] This has exactly one solution in \[ \left(-\frac{\pi}{4},\frac{\pi}{4}\right). \]
Step 4: Conclusion
Hence there is exactly one real root.
Final Answer: (B)
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