A) \[\frac{\pi }{3}\left( 6m-1 \right)\]
B) \[\frac{\pi }{3}\left( 6m+1 \right)\]
C) \[\frac{\pi }{3}\left( 2m+1 \right)\]
D) none of these (Where \[m\in I\])
Correct Answer: B
Solution :
| \[1+\cos 3x+1-\left[ \cos \left( \frac{\pi }{2}+\left( 2x-\frac{7\pi }{6} \right) \right) \right]=0\] |
| \[2{{\cos }^{2}}\frac{3x}{2}+1-\cos \left( 2x-\frac{2\pi }{3} \right)=0\] |
| \[2{{\cos }^{2}}\frac{3x}{2}+2{{\sin }^{2}}\left( x-\frac{\pi }{3} \right)=0\] |
| \[\cos \frac{3x}{2}=0,\sin \left( x-\frac{\pi }{3} \right)=0\] |
| \[x=\frac{\pi }{3},\pi \] and \[x=\frac{\pi }{3},\frac{2\pi }{3},\frac{7\pi }{3},....\] |
| \[\therefore \]\[x=\frac{\pi }{3}\]is the common value which satisfies both |
| \[\therefore \]\[x=2m\pi +\frac{\pi }{3}=(6m+1)\frac{\pi }{3}.\] |
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