A) \[4\]
B) \[6\]
C) \[5\]
D) \[2\]
Correct Answer: B
Solution :
\[\sin x+\sin 5x=\sin 3x\] \[\Rightarrow \] \[2\sin \left( \frac{x+5x}{2} \right).\cos \left( \frac{5x-x}{2} \right)=\sin 3x\] \[\Rightarrow \] \[2\sin 3x.cos\,2x=sin3x\] \[\Rightarrow \] \[\sin 3x\,(2\cos \,2x-1)=0\] \[\Rightarrow \] \[\sin 3x=0\] and \[2\cos 2x-1=0\] \[\Rightarrow \] \[\sin 3x=sin0\] and \[\cos 2x=\frac{1}{2}=\cos \frac{\pi }{3}\] \[\Rightarrow \] \[3x=n\pi \] and \[2x=2n\pi \pm \frac{\pi }{3}\] \[\Rightarrow \] \[x=\frac{n\pi }{3}\] and \[x=n\pi \pm \frac{\pi }{6}\] where \[n\in z,\] \[\Rightarrow \] \[x=0,\,\frac{\pi }{3},\frac{2\pi }{3},\frac{3\pi }{3}\] and \[x=\frac{\pi }{6},\frac{5\pi }{6}\] For \[n=0,1,2,3....\] So, the total number of solution lie between \[[0,\,\pi ]\] \[x=0,\frac{\pi }{6},\frac{\pi }{3},\frac{2\pi }{3},\frac{5\pi }{6},\frac{3\pi }{3}\]
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