Find the general solution for $\cos 4x = \cos 2x$
A. $x = n\pi $ or $\dfrac{{n\pi }}{6}$
B. $x = n\pi $ or $\dfrac{{n\pi }}{3}$
C. $x = \dfrac{{2n\pi }}{3}$
D. $x = \pi $

Hint: To solve this question, we will use the concept of trigonometric equations. Equations involving a variable's trigonometric functions are called trigonometric equations. The general solution of a trigonometric equation can be identified by using the theorem: For any real numbers x and y, $\cos x = \cos y$, implies $x = 2n\pi \pm y,{\text{ where n}} \in {\text{Z}}$

Complete step-by-step answer:
The solutions of a trigonometric equation for which $0 \leqslant x < 2\pi $ are called principal solutions.
The general solution is called the expression involving integer 'n' which gives all solutions of a trigonometric equation.
Given that,
$\cos 4x = \cos 2x$
This can also be written as:
$\cos 4x - \cos 2x = 0$ ………. (i)
As we know that,
$\cos x - \cos y = - 2\sin \dfrac{{x + y}}{2}\sin \dfrac{{x - y}}{2}$
So, the equation (i) will become,
$\cos 4x - \cos 2x = - 2\sin \left( {\dfrac{{4x + 2x}}{2}} \right)\sin \left( {\dfrac{{4x - 2x}}{2}} \right)$
$\
   \Rightarrow - 2\sin \left( {\dfrac{{6x}}{2}} \right)\sin \left( {\dfrac{{2x}}{2}} \right) \\
   \Rightarrow - 2\sin 3x\sin x \\
$
Putting this value in equation (i), we will get
$ \Rightarrow - 2\sin 3x\sin x = 0$
Here, we can say that either $\sin 3x = 0$ or $\sin x = 0$
It has been observed that if x increases (or decreases) by any integral multiple of $2\pi $, the values of sine functions do not change.
Thus,
$\sin \left( {2n\pi + x} \right) = \sin x,n \in Z$
Further $\sin x = 0$, if $x = 0, \pm \pi , \pm 2\pi \pm 3\pi ,........,$ i.e. when x is an integral multiple of $\pi $
Thus,
$\sin x = 0$ implies $x = n\pi $, where n is any integer.
So, we will solve $\sin 3x = 0$ and $\sin x = 0$ separately.
1. General solution for $\sin 3x = 0$
We know that,
If $\sin x = 0$, then
$x = n\pi $
Therefore,
$\sin 3x = 0$ implies $3x = n\pi $
We will get,
$x = \dfrac{{n\pi }}{3}$
2. General solution for $\sin x = 0$
If $\sin x = 0$, implies
$x = n\pi $
Hence, we can say that the general solutions of $\cos 4x = \cos 2x$ are $x = n\pi $ or $\dfrac{{n\pi }}{3}$
Therefore, the correct answer is option (B).

Note:Whenever we ask such types of questions, we have to remember some basic points to solve a trigonometric equation. First, we have to make a trigonometric equation that is equals to 0. Then we will simplify that equation in terms of trigonometric functions. After that we will put that simplified equation to 0 and we will get some cases. Then we will find out the general solutions for those cases and through this, we will get the required answer.