The general solution of \[\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x\] is
A.\[n\pi +\dfrac{\pi }{8}\]
B.\[\dfrac{n\pi }{2}+\dfrac{\pi }{8}\]
C.\[{{\left( -1 \right)}^{n}}\left( \dfrac{n\pi }{2}+\dfrac{\pi }{8} \right)\]
D.\[2n\pi -\dfrac{\pi }{8}\]

Hint: We know the formula, \[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] and \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] . Use this formula and simplify \[\sin x+\sin 3x\] and \[\cos x+\cos 3x\] and then, substitute it in the equation \[\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x\] . We know that the maximum value of cosine function is 1. We know that \[\tan \dfrac{\pi }{4}=1\] and the general solution of \[\tan x=\tan \dfrac{\pi }{4}\] is \[x=n\pi +\dfrac{\pi }{4}\] .

Complete step-by-step answer:
According to the question, we have to find the general solution of,
\[\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x\] ……………..(1)
We know the formula,
\[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] …………………(2)
Replacing A by x and B by 3x in the equation (2), we get
\[\sin x+\sin 3x=2\sin \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)\]
\[\Rightarrow \sin x+\sin 3x=2\sin 2x\cos (-x)\] ……………….(4)
We know the property that, \[\cos (-x)=\cos x\] .
Using this property in equation (4), we get
\[\Rightarrow \sin x+\sin 3x=2\sin 2x\cos (-x)\]
\[\Rightarrow \sin x+\sin 3x=2\sin 2x\cos x\] …………………….(5)
We also know the formula,
\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] …………………(6)
Replacing A by x and B by 3x in the equation (6), we get
\[\cos x+\cos 3x=2\cos \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right)\]
\[\Rightarrow \cos x+\cos 3x=2\cos 2x\cos (-x)\] ……………….(7)
We know the property that, \[\cos (-x)=\cos x\] .
Using this property in equation (7), we get
\[\Rightarrow \cos x+\cos 3x=2\cos 2x\cos (-x)\]
\[\Rightarrow \cos x+\cos 3x=2\cos 2x\cos x\] …………………….(8)
From equation (1), we have
\[\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x\]
\[\Rightarrow \sin x+\sin 3x-3\sin 2x=\cos x+\cos 2x-3\cos 2x\] …………………(9)
Now, from equation (5), equation (8), and equation (9), we get
\[\Rightarrow \sin x+\sin 3x-3\sin 2x=\cos x+\cos 3x-3\cos 2x\]
\[\Rightarrow 2\sin 2x\cos x-3\sin 2x=2\cos 2x\cos x-3\cos 2x\]
Taking \[\sin 2x\] in the above equation, we get
\[\Rightarrow \sin 2x\left( 2\cos x-3 \right)=\cos 2x\left( 2\cos x-3 \right)\]
\[\begin{align}
  & \Rightarrow \sin 2x\left( 2\cos x-3 \right)-\cos 2x\left( 2\cos x-3 \right)=0 \\
 & \Rightarrow \left( 2\cos x-3 \right)\left( \sin 2x-\cos 2x \right)=0 \\
\end{align}\]
\[\left( 2\cos x-3 \right)=0\] or \[\left( \sin 2x-\cos 2x \right)=0\] .
First, let us solve for \[\left( 2\cos x-3 \right)=0\]
\[\left( 2\cos x-3 \right)=0\]
\[\begin{align}
  & \Rightarrow 2\cos x=3 \\
 & \Rightarrow \cos x=\dfrac{3}{2} \\
\end{align}\]
We know that the maximum value of the cosine function is 1 and here, we have \[\cos x=\dfrac{3}{2}\] , which is not possible for any value of x.
Now, we have to solve \[\left( \sin 2x-\cos 2x \right)=0\] .
\[\left( \sin 2x-\cos 2x \right)=0\]
\[\Rightarrow \sin 2x=\cos 2x\]
\[\Rightarrow \dfrac{\sin 2x}{\cos 2x}=1\] …………….(10)
We know that, \[\dfrac{\sin A}{\cos A}=\tan A\] .
Replacing A by 2x in \[\dfrac{\sin A}{\cos A}=\tan A\] , we get
\[\dfrac{\sin 2x}{\cos 2x}=\tan 2x\] ………(11)
From equation (10) and equation (11), we get
\[\Rightarrow \dfrac{\sin 2x}{\cos 2x}=1\]
\[\Rightarrow \tan 2x=1\] …………..(12)
Using \[\tan \dfrac{\pi }{4}=1\] in equation (12), we get
\[\begin{align}
  & \Rightarrow \tan 2x=\tan \dfrac{\pi }{4} \\
 & \Rightarrow 2x=\dfrac{\pi }{4} \\
\end{align}\]
Using \[\tan \dfrac{5\pi }{4}=1\] in equation (12), we get
\[\begin{align}
  & \Rightarrow \tan 2x=\tan \dfrac{5\pi }{4} \\
 & \Rightarrow 2x=\dfrac{5\pi }{4} \\
\end{align}\]
We have \[2x=\dfrac{\pi }{4}\] , \[2x=\dfrac{5\pi }{4}=\pi +\dfrac{\pi }{4}\] ……………….
Now, in general we can say that,
\[\begin{align}
  & 2x=n\pi +\dfrac{\pi }{4} \\
 & \Rightarrow x=\dfrac{n\pi }{2}+\dfrac{\pi }{8} \\
\end{align}\]
So, the correct option is option (B).

Note: In this question, one might think to find the general solution of \[\cos x=\dfrac{3}{2}\] . But we don’t have any solution for \[\cos x=\dfrac{3}{2}\] because the maximum value of cosine function is 1. Also, one can think to use formula \[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\] and \[\sin 2x=2\sin x\cos x\] in the equation
\[\sin x-3\sin 2x+\sin 3x=\cos x-3\cos 2x+\cos 3x\] . If we do so, then our equation will become complex to solve.