A) \[n\pi +\frac{\pi }{8}\]
B) \[\frac{n\pi }{2}+\frac{\pi }{8}\]
C) \[{{(-1)}^{n}}\frac{n\pi }{2}+\frac{\pi }{8}\]
D) \[2n\pi +{{\cos }^{-1}}\frac{3}{2}\]
Correct Answer: B
Solution :
\[\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x\] \[\Rightarrow \]\[2\sin 2x\cos x-3\sin 2x-2\cos 2x\cos x+3\cos 2x=0\] \[\Rightarrow \] \[\sin 2x(2\cos x-3)-\cos 2x(2\cos x-3)=0\] \[\Rightarrow \]\[(\sin 2x-\cos 2x)\,\,(2\cos x-3)=0\Rightarrow \sin 2x=\cos 2x\] \[\Rightarrow \] \[2x=2n\pi \pm \left( \frac{\pi }{2}-2x \right)\] i.e., \[x=\frac{n\pi }{2}+\frac{\pi }{8}\].
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