A) 0
B) 2
C) 4
D) 6
Correct Answer: C
Solution :
\[\frac{1}{\sin x}-\frac{1}{\sin 2x}=\frac{2}{\sin 4x}\] \[\Rightarrow \] \[\frac{\sin 2x-\sin x}{\sin x\sin 2x}=\frac{2}{2\sin 2x\,\cos 2x}\] \[\Rightarrow \] \[2\sin 2x\,\cos 2x-2\,\sin x\,\cos 2x=2\,\sin x\] \[\Rightarrow \] \[\sin 4x-\sin 3x+\sin x=2\,\sin x\] \[\Rightarrow \] \[\sin 4x=\sin 3x+\sin x\] \[\Rightarrow \] \[2\sin 2x\cos 2x=2\sin 2x\,\cos x\] \[\Rightarrow \] \[2x=2n\pi \pm x\,(\text{as}\,\sin \,2x\ne 0)\] \[\Rightarrow \] \[x=\frac{2n\pi }{3},\]\[n\in I\](as\[x=2k\]is not in domain) \[n=1,2,4,5,...\] \[\therefore \] Four solutions are\[\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}\].
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