Answer
The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{8}$, $\frac{7\pi }{8}$, $\frac{9\pi }{8}$ and $\frac{15\pi }{8}.$
Work Step by Step
We know that the period of the cosine function is $2\pi $. In the interval there are two values at which the cosine function is $\frac{\sqrt{2}}{2}$. One is $\frac{\pi }{8}$ and since cosine is positive in quadrant IV, thus the other value is:
$\begin{align}
& 2\pi -\frac{\pi }{4}=\frac{8\pi -\pi }{4} \\
& =\frac{7\pi }{4}
\end{align}$
So, all the solutions to $\cos 2x=\frac{\sqrt{2}}{2}$ are given by:
$\begin{align}
& 2x=\frac{\pi }{4}+2n\pi \\
& x=\frac{\pi }{8}+n\pi
\end{align}$
Or,
$\begin{align}
& 2x=\frac{7\pi }{4}+2n\pi \\
& x=\frac{7\pi }{8}+n\pi
\end{align}$
Where, n is any integer. and the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$ and $n=1$. And the equation is calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=\frac{\pi }{8}+n\pi \\
& =\frac{\pi }{8}+0\times \pi \\
& =\frac{\pi }{8}+0 \\
& =\frac{\pi }{8}
\end{align}$
$\begin{align}
& x=\frac{\pi }{8}+n\pi \\
& =\frac{\pi }{8}+1\times \pi \\
& =\frac{\pi +8\pi }{8} \\
& =\frac{9\pi }{8}
\end{align}$
The second equation is also evaluated by taking first $n$ as 0 and then as 1. It can be further simplified as follows.
$\begin{align}
& x=\frac{7\pi }{8}+n\pi \\
& =\frac{7\pi }{8}+0\times \pi \\
& =\frac{7\pi }{8}+0 \\
& =\frac{7\pi }{8}
\end{align}$
$\begin{align}
& x=\frac{7\pi }{8}+n\pi \\
& =\frac{7\pi }{8}+1\times \pi \\
& =\frac{7\pi +8\pi }{8} \\
& =\frac{15\pi }{8}
\end{align}$
