A) \[\frac{\pi }{3}(6k+1)\]
B) \[\frac{\pi }{3}(6k+1)\]
C) \[\frac{\pi }{3}(2k+1)\]
D) None of these
Correct Answer: A
Solution :
We have \[\cos 3x+\sin \text{ }\left( 2x-\frac{7\pi }{6} \right)\,=-2\] \[\Rightarrow \] \[1+\cos 3x+1+\sin \left( 2x-\frac{7\pi }{6} \right)=0\] \[\Rightarrow \] \[(1+\cos 3x)+1-\cos \left( 2x-\frac{2\pi }{3} \right)=0\] \[\Rightarrow \] \[2{{\cos }^{2}}\frac{3x}{2}+2{{\sin }^{2}}\left( x-\frac{\pi }{3} \right)=0\] \[\Rightarrow \] \[\cos \frac{3x}{2}=0\] and \[\sin \left( x-\frac{\pi }{3} \right)=0\] \[\Rightarrow \]\[\frac{3x}{2}=\frac{\pi }{2},\,\frac{3\pi }{2},\,.....\]and \[x-\frac{\pi }{3}\]=0, \[\pi ,2\pi .....\Rightarrow x=\frac{\pi }{3}\] Therefore, the general solution of \[\cos \frac{3x}{2}=0\] and \[\sin \left( x-\frac{\pi }{3} \right)=0\]is\[x=2k\pi +\frac{\pi }{3}=\frac{\pi }{3}(6k+1),\]where \[k\in Z\].
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