A) \[k\pi ,k\in I\]
B) \[2k\pi ,k\in I\]
C) \[k\frac{\pi }{2},k\in I\]
D) None of these
Correct Answer: B
Solution :
We have, \[1-\cos \theta =\sin \theta \,.\,\sin \frac{\theta }{2}\] Þ \[2{{\sin }^{2}}\frac{\theta }{2}=2\sin \frac{\theta }{2}\,.\,\cos \frac{\theta }{2}\,.\,\sin \frac{\theta }{2}\] Þ \[2{{\sin }^{2}}\frac{\theta }{2}\,\left[ 1-\cos \frac{\theta }{2} \right]=0\]Þ \[\sin \frac{\theta }{2}=0\]or\[2{{\sin }^{2}}\frac{\theta }{4}=0\] Þ \[\sin \frac{\theta }{2}\] = 0 or \[\sin \frac{\theta }{4}=0\]Þ \[\frac{\theta }{2}=k\pi \] or\[\frac{\theta }{4}=k\pi \]. Hence, \[\theta =2k\pi \] or\[\theta =4k\pi \], \[k\in I\].
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