A) \[3\]
B) \[4\]
C) \[5\]
D) \[6\]
Correct Answer: C
Solution :
Given equation is \[\sin x\cos 3x=\sin 3x\,\cos \,5x\] \[\Rightarrow \] \[2\sin x\,\cos \,3x-2\sin 3x\,cos5x=0\] \[\Rightarrow \] \[\sin (3x+x)-\sin (3x-x)-\sin (3x+5x)\] \[+\sin (5x-3x)=0\] \[\Rightarrow \] \[\sin 4x-\sin 2x-\sin 8x+\sin 2x=0\] \[\Rightarrow \] \[\sin 4x-\sin 8x=0\] \[\Rightarrow \] \[2\cos \left( \frac{4x+8x}{2} \right)\sin \left( \frac{8x-4x}{2} \right)=0\] \[\Rightarrow \] \[2\cos \,6x\,\sin \,2x=0\] \[\Rightarrow \] \[\cos \,\,6x=0\] or \[\sin \,2x=0\] \[\Rightarrow \] \[6x=(2n+1)\frac{\pi }{2}\] or \[x=\frac{n\pi }{2}\] \[\Rightarrow \] \[x=(2n+1)\frac{\pi }{12}\] or \[x=\frac{n\pi }{2}\] \[\Rightarrow \] \[x=0,\,\frac{\pi }{2},\frac{\pi }{12},\frac{3\pi }{12},\frac{5\pi }{12}\in \left[ 0,\frac{\pi }{2} \right]\] \[\therefore \] Number of solutions is 5.
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