A) \[\frac{n\pi }{3}\]
B) \[n\pi +\frac{\pi }{3}\]
C) \[2n\pi \pm \frac{\pi }{3}\]
D) None of these
Correct Answer: A
Solution :
\[2\sin 3x\cos x-2\sin 3x=0\], \[\therefore \] \[\sin 3x=0\], \[\cos x=1\] \[\frac{1}{2}=\frac{\frac{3h}{120}-\frac{h}{120}}{1+\frac{3{{h}^{2}}}{14400}}\Rightarrow h=120,\,40\] \[3x=n\pi \] or \[x=\frac{n\pi }{3}\] and \[x=2n\pi \] The second value \[x=2n\pi \] is included in the value given by\[x=\frac{n\pi }{3}\].
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