A) \[n\pi +\frac{\pi }{2}\]
B) \[n\pi +\frac{\pi }{4},\,\,n\pi \]
C) \[n\pi +\frac{\pi }{2}\]
D) \[n\pi +\frac{\pi }{4},\,\,\frac{\pi }{4}\]
Correct Answer: B
Solution :
The given system has a non-trivial solution, if \[\left| \begin{matrix} \lambda & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ -1 & \sin \alpha & -\cos \alpha \\ \end{matrix} \right|=0\] By expanding the determinant along\[({{C}_{1}})\], we get \[\lambda (-{{\cos }^{2}}ga-{{\sin }^{2}}\alpha )-1(-\sin \alpha \cos \alpha \] \[-\sin \alpha \cos \alpha )\] \[-1({{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha )=0\] For \[\lambda =1,\,\,\sin 2\alpha +\cos 2\alpha =1\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}\sin 2\alpha +\frac{1}{\sqrt{2}}\cos 2\alpha =\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[\cos \left( 2\alpha -\frac{\pi }{4} \right)=\cos \left( 2n\pi \pm \frac{\pi }{4} \right)\] \[\Rightarrow \]\[2\alpha =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{4},\,\,n\]being an integer. \[\Rightarrow \] \[\alpha =n\pi +\frac{\pi }{4},\,\,n\pi \]
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