## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{8}$, $\frac{7\pi }{8}$, $\frac{9\pi }{8}$ and $\frac{15\pi }{8}.$
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}