Question
The general solution of $$\sin \,x - 3\,\sin \,2x\, + \sin \,3x\, = \cos x - 3\,\cos \,\,2x + \cos \,3x$$ is
A.
$$n\pi + \frac{\pi }{8}$$
B.
$$\frac{{n\pi }}{2} + \frac{\pi }{8}$$
C.
$${\left( { - 1} \right)^n}\frac{{n\pi }}{2} + \frac{\pi }{8}$$
D.
$$2n\pi + {\cos ^{ - 1}}\frac{3}{2}$$
Answer :
$$\frac{{n\pi }}{2} + \frac{\pi }{8}$$
Solution :
The given equation is
$$\eqalign{
& \sin \,x - 3\,\sin \,2x\, + \sin \,3x\, = \cos x - 3\,\cos \,\,2x + \cos \,3x \cr
& \Rightarrow \,2\,\sin \,2x\,\cos \,x - 3\,\sin \,2x = 2\,\cos \,\,2x\,\cos \,x - 3\,\cos \,\,2x \cr
& \Rightarrow \,\sin \,2x\,\left( {2\,\cos \,x - 3} \right) = \cos \,\,2x\,\left( {2\,\cos \,x - 3} \right) \cr
& \Rightarrow \,\sin \,2x = \cos \,2x\,\,\,\,\left( {{\text{as}}\,\cos \,x\, \ne \,\frac{3}{2}} \right) \cr
& \Rightarrow \,\tan \,2x\, = 1\, \cr
& \Rightarrow \,2x = n\,\pi \,\, + \,\frac{\pi }{4} \cr
& \Rightarrow \,x = \frac{{n\pi }}{2} + \frac{\pi }{8} \cr} $$