A) 0
B) 2
C) 1
D) 3
Correct Answer: C
Solution :
Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\]and taking common from \[{{R}_{1}},\]we get \[(\sin x+2\cos x)\left| \begin{matrix} 1 & 1 & 1 \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(\sin x+2\cos x)\] \[\Rightarrow \]\[\left| \begin{matrix} 1 & 0 & 0 \\ \cos x & \sin x-\cos x & 0 \\ \cos x & 0 & \sin x-\cos x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[{{(\sin \,x-\cos x)}^{2}}(\sin x+2\cos x)=0\] \[\Rightarrow \] \[x=\frac{\pi }{4},x\in \left[ -\frac{\pi }{4},\frac{\pi }{4} \right]\]
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