If $\cos 2 x=(\sqrt{2}+1)(\cos x-(1 / \sqrt{2})), \cos x \neq 1 / 2, x \in I$
1) $\{2 n \pi \pm \pi / 3: n \in Z\}$
2) $\{2 n \pi \pm \pi / 6: n \in Z\}$
3) $\{2 n \pi \pm \pi / 2: n \in Z\}$
4) $\{2 n \pi \pm \pi / 4: n \in Z\}$

Hint: In this question, we are given a trigonometric equation. So first we convert the cos2x in another trigonometric equation and by solving further we find the value of cosx. And then by using trigonometric functions, we find the value of x and then its value.

Formula Used:
 $\cos 2x=2\cos 2x-1$

Complete step by step Solution:
Given that
$\cos (2 x)=(\sqrt{2}+1)\left(\cos (x)-\left(\dfrac{1}{\sqrt{2}}\right)\right)$
We know $\cos 2x=2\cos 2x-1$
And then multiply and divide the R.H.S by $\sqrt{2}$
 the given equation becomes
$2\cos 2x-1=\dfrac{\sqrt{2}+1}{\sqrt{2}}\left( \sqrt{2}\cos x-1 \right)$
$(\sqrt{2}\cos x+1)(\sqrt{2}\cos x-1)=\dfrac{\sqrt{2}+1}{\sqrt{2}}(\sqrt{2}\cos x-1)$
Then $(\sqrt{2}\cos x+1)=\dfrac{\sqrt{2}+1}{\sqrt{2}}$
$\sqrt{2}\cos x=1+\dfrac{1}{\sqrt{2}}-1$
Hence $\sqrt{2}\cos x=\dfrac{1}{\sqrt{2}}$
And $\cos x=\dfrac{1}{\sqrt{2}\times \sqrt{2}}$
Then $\cos x=\dfrac{1}{2}$
The algebraic approach to solving simultaneous linear equations is known as the substitution method. This approach, as the name suggests, involves substituting the value of a variable based on one equation into the second equation.
By doing this, a pair of linear equations are combined into one linear equation with just one variable, making it much simpler to solve.
$\cos (x)=\dfrac{1}{2} \quad: \quad x=\dfrac{\pi}{3}+2 \pi n, x=\dfrac{5 \pi}{3}+2 \pi n$
Therefore the required value is $\cos (x)=\dfrac{1}{2} \quad: \quad x=\dfrac{\pi}{3}+2 \pi n, x=\dfrac{5 \pi}{3}+2 \pi n$

Hence, the correct option is 1.

Additional information:
Trigonometric Identities are equality statements that hold true for all values of the variables in the equation and that use trigonometry functions.
All trigonometric identities are built upon the foundation of the six trigonometric ratios. Some of their names are sine, cosine, tangent, cosecant, secant, and cotangent. Each of these trigonometric ratios is defined using the adjacent side, opposite side, and hyperbolic tangent side of the right triangle. All basic trigonometric identities are derived from the six trigonometric ratios.

Note: In these types of questions, Students made mistakes in finding the method to solve the equations. These types of questions need practice so that our minds will be able to find the easy method to solve the question. We must remember the trigonometric equations to find out the answer.