## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{4}$, $\frac{\pi }{2}$, $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$.
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} $=\frac{3\pi }{2}$