Answer
The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{4}$, $\frac{\pi }{2}$, $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$.
Work Step by Step
We know that the period of the sine function is $2\pi $. In the interval there are two values at which the sine function is $\frac{\sqrt{2}}{2}$. One is $\frac{\pi }{4}$ . And the sine is positive in quadrant II, thus the other value is:
$\begin{align}
& \pi -\frac{\pi }{4}=\frac{4\pi -\pi }{4} \\
& =\frac{3\pi }{4}
\end{align}$
Therefore, all the solutions to $\sin \left( 2x-\frac{\pi }{4} \right)=\frac{\sqrt{2}}{2}$ be given by:
$\begin{align}
& 2x-\frac{\pi }{4}=\frac{\pi }{4}+2n\pi \\
& 2x=\frac{2\pi }{4}+2n\pi \\
& \,x=\frac{\pi }{4}+n\pi
\end{align}$
Or,
$\begin{align}
& 2x-\frac{\pi }{4}=\frac{3\pi }{4}+2n\pi \\
& 2x=\frac{4\pi }{4}+2n\pi \\
& x=\frac{\pi }{2}+n\pi
\end{align}$
Where n is any integer. And the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$, $n=1$. And the equation is calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=\frac{\pi }{4}+n\pi \\
& =\frac{\pi }{4}+0\times \pi \\
& =\frac{\pi }{4}+0 \\
& =\frac{\pi }{4}
\end{align}$
$\begin{align}
& x=\frac{\pi }{4}+n\pi \\
& =\frac{\pi }{4}+1\times \pi \\
& =\frac{\pi }{4}+1\pi \\
& =\frac{\pi +4\pi }{4}
\end{align}$
$=\frac{5\pi }{4}$
The second equation is also computed by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=\frac{\pi }{2}+n\pi \\
& =\frac{\pi }{2}+0\times \pi \\
& =\frac{\pi }{2}+0 \\
& =\frac{\pi }{2}
\end{align}$
$\begin{align}
& x=\frac{\pi }{2}+n\pi \\
& =\frac{\pi }{2}+1\times \pi \\
& =\frac{\pi }{2}+1\pi \\
& =\frac{\pi +2\pi }{2}
\end{align}$
$=\frac{3\pi }{2}$
