Please learn to post the complete question. You seem to add that there are limitations on the domain?
Just look at $\cos \left( {2x - \frac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}$
It is easily seen that $2x - \frac{\pi }{4}=\pm\dfrac{\pi}{4}+2k\pi$
Note that $x=0$ is also a solution.
Now you finish.
Sorry. The purpose of this question is to solve for x within the domain of [0, 2pi)
I will provide a specific example to explain where I am confused. So once you find the first solution for 2x-(pi/4), which is pi/4, I'm wondering what the steps are you take in order to manipulate that value to produce the value in terms of just x? Do you multiply pi/2 by 1/2 and then add pi/4 or do you add pi/4 first and then multiply by 1/2?
Zero is a solution. Recall that $\cos$ is an even function:
$\cos \left( {2 \times 0 - \frac{\pi }{4}} \right) = \underbrace {\cos \left( {\frac{-\pi }{4}} \right)}_{even}= \cos \left( { \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}$
$\pi$ is also a solution: $2\times\pi-\frac{\pi}{4}=\frac{7\pi}{4}$ and $\cos\left(\frac{7\pi}{4}\right)=\frac{\sqrt2}{2}$ .
$\cos \left(2x - \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$
from the unit circle, cosine of an angle equals $\dfrac{\sqrt{2}}{2}$ at an angle of $\dfrac{\pi}{4}$ above and below the positive x-axis (quadrants I and IV)
$0 \le x < 2\pi \implies -\dfrac{\pi}{4} \le 2x-\dfrac{\pi}{4} < \dfrac{15\pi}{4}$
$2x-\dfrac{\pi}{4} = \bigg\{-\dfrac{\pi}{4}, \, \dfrac{\pi}{4},\, \dfrac{7\pi}{4}, \, \dfrac{9\pi}{4} \bigg\}$
$2x = \bigg\{0, \, \dfrac{\pi}{2},\, 2\pi, \, \dfrac{5\pi}{2} \bigg\}$
$x = \bigg\{0, \, \dfrac{\pi}{4},\, \pi, \, \dfrac{5\pi}{4} \bigg\}$