A) \[x=(2n+1)\frac{\pi }{7},\,n\in Z\]
B) \[x=(2n+1)\frac{\pi }{5},\,n\in Z\]
C) \[x=(2n+1)\frac{\pi }{35},\,n\in Z\]
D) \[x=(2n+1)\,\pi ,\,n\in Z\]
Correct Answer: D
Solution :
Given, \[\cos x.\,\cos \,6x=-1\] \[\Rightarrow \] \[\cos \,x=-1\] and \[\cos \,\,6x=-1\] \[\Rightarrow \] \[\cos \,x=\cos \pi \] and \[\cos 6x=\cos \pi \] \[\Rightarrow \] \[x=2n\pi \pm \pi \] and \[\cos \,6x=\cos \pi \] \[\Rightarrow \] \[x=(2n\pm 1)\pi \] and \[6x=2n\pi \pm \pi \] \[x=(2n\pm 1)\frac{\pi }{6}\] where \[n\in z,\]
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