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 :
\[\sin \,x -3 sin2x\,+ sin 3x = cos\,x\,-3 cos\,2x\,+ cos 3x\]\[\Rightarrow \,\,\,2sin\,2x\,cox\,x-3sin2x=2cos2xcos\,x-3cos2x\] \[\Rightarrow \,\,\sin 2x \left( 2 cos\,x-3 \right) = cos2x\left( 2 cos\,x- 3 \right)\] \[\Rightarrow \,\,\sin 2x =\cos 2x\,\left[ at\,cos\,x\ne \frac{3}{2} \right]\]
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