A) \[2n\pi \pm \frac{\pi }{4},n\in Z\]
B) \[2n\pi \pm \frac{\pi }{3},n\in Z\]
C) \[2n\pi \pm \frac{\pi }{6},n\in Z\]
D) \[2n\pi \pm \frac{\pi }{2},n\in Z\]
Correct Answer: A
Solution :
\[\cos \,\,x+\cos \,3x\,+cos\,2x=0\] \[\Rightarrow \] \[2\,\cos \,2x\cos x+\cos \,2x=0\] \[\Rightarrow \] \[\cos \,2x(2\,cos\,x+1)=0\] \[\Rightarrow \] \[\cos \,\,2x=0\,\,\,\,\left( \because \,\,\cos \,x\ne -\frac{1}{2} \right)\] \[\Rightarrow \] \[2x=\frac{\pi }{2}\] \[\Rightarrow \] \[x=\frac{\pi }{4}\] \[\therefore \] General value is \[2n\pi \pm \frac{\pi }{4},\,\,n\in Z\]
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