If we have the trigonometric equation as \[\cos 2x=\left( \sqrt{2}+1 \right)\left( \cos x-\left( \dfrac{1}{\sqrt{2}} \right) \right),\cos x\ne \dfrac{1}{2},x\in I\] then find the solution.
1.\[\left\{ 2n\pi \pm \dfrac{\pi }{3}:n\in Z \right\}\]
2.\[\left\{ 2n\pi \pm \dfrac{\pi }{6}:n\in Z \right\}\]
3.\[\left\{ 2n\pi \pm \dfrac{\pi }{2}:n\in Z \right\}\]
4.\[\left\{ 2n\pi \pm \dfrac{\pi }{4}:n\in Z \right\}\]

Hint: In order to solve it, we will be considering the given expression. We will be solving both the LHS and RHS simultaneously. Then we will be trying to expand the LHS term conveniently so that we would get common terms on both sides. We will be solving it in such a way that we would be obtaining the value of \[\cos x\]. Then we will be obtaining the angular value of \[x\] in terms of principal angles.

Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between \[{{0}^{\circ }}\] and \[{{360}^{\circ }}\]. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let us start solving the given problem.
We are given with \[\cos 2x=\left( \sqrt{2}+1 \right)\left( \cos x-\left( \dfrac{1}{\sqrt{2}} \right) \right)\]
Now let us solve this accordingly, we get
\[\begin{align}
  & \cos 2x=\left( \sqrt{2}+1 \right)\left( \cos x-\left( \dfrac{1}{\sqrt{2}} \right) \right) \\
 & \Rightarrow 2{\cos}^2x-1=\left( \dfrac{\sqrt{2}+1}{\sqrt{2}} \right)\left( \sqrt{2}\cos x-1 \right) \\
\end{align}\]
On further solving,
\[\begin{align}
  & \Rightarrow 2{\cos}^2{x}-1=\left( \dfrac{\sqrt{2}+1}{\sqrt{2}} \right)\left( \sqrt{2}\cos x-1 \right) \\
 & \Rightarrow \left( \sqrt{2}\cos x+1 \right)\left( \sqrt{2}\cos x-1 \right)=\left( \dfrac{\sqrt{2}+1}{\sqrt{2}} \right)\left( \sqrt{2}\cos x-1 \right) \\
\end{align}\]
Upon bringing the term \[\left( \sqrt{2}\cos x-1 \right)\] from RHS to LHS and upon solving it, we get
\[\begin{align}
  & \Rightarrow \left( \sqrt{2}\cos x+1 \right)\left( \sqrt{2}\cos x-1 \right)=\left( \dfrac{\sqrt{2}+1}{\sqrt{2}} \right)\left( \sqrt{2}\cos x-1 \right) \\
 & \Rightarrow \dfrac{\left( \sqrt{2}\cos x+1 \right)\left( \sqrt{2}\cos x-1 \right)}{\left( \sqrt{2}\cos x-1 \right)}=\left( \dfrac{\sqrt{2}+1}{\sqrt{2}} \right) \\
 & \Rightarrow \left( \sqrt{2}\cos x+1 \right)=\left( \dfrac{\sqrt{2}+1}{\sqrt{2}} \right) \\
\end{align}\]
Now we will be transposing 1 from LHS to the RHS and let us split the terms in RHS. We get
\[\begin{align}
  & \Rightarrow \sqrt{2}\cos x=1+\dfrac{1}{\sqrt{2}}-1 \\
 & \Rightarrow \sqrt{2}\cos x=\dfrac{1}{\sqrt{2}} \\
 & \Rightarrow \cos x=\dfrac{1}{2} \\
\end{align}\]
We have obtained the value \[\cos x=\dfrac{1}{2}\], but we are given that \[\cos x\ne \dfrac{1}{2}\]. So we will be considering the angle \[\left\{ 2n\pi \pm \dfrac{\pi }{4}:n\in Z \right\}\]
Hence option 4 is the correct answer.

Note: We must always try to express the angles given in terms of principal angles as it is the formal way of expressing. While expanding the terms, we must expand the trigonometric terms correctly. Else, the whole expansion would be incorrect. This is the most commonly committed error. We must be aware of all the possible expansions.