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} , \frac{\pi }{4} ] $ is
Answer & explanation
Correct answer: option 2
The correct answer is option (2) : -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 $
$⇒\begin{vmatrix} sin\, x+2\, cos \, x & sin\, x+2\, cos \, x & sin\, x+2\, cos \, x \\cos\, x & sin \, x & cos \, x\\ cos\, x & cos \, x& sin \, x \end{vmatrix}=0$
[Applying $R_1→R_1+R_2+R_3 $ ]
$⇒(sin\, 2x+ 2cos \, x) \begin{vmatrix} 1 & 1& 1\\cos\, x & sin \, x & cos \, x\\ cos\, x & cos \, x& sin \, x \end{vmatrix}=0$
$⇒(sin\, x + 2\, cos \, x) \begin{vmatrix} 1 & 0& 0\\cos\, x & sin \, x-cos\, x & 0\\ cos\, x & 0 & sin \, x-cos\, x \end{vmatrix}=0$
[Applying $C_3→C_1, C_2→C_2-C_1$]
$⇒(sin \, x + 2cos\, x) (sin \, x -cos\, x)^2 =0$
$⇒tan\, x =-2\, or\, tan\, x = 1$
$⇒x=\frac{\pi}{4} $ $[∵x \in [-\frac{\pi }{4}, \frac{\pi }{4} ]⇒-1≤tan\, x≤1]$
Hence, there is only one real root in $[-\frac{\pi }{4} , \frac{\pi }{4} ]$