## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{6}$, $\frac{\pi }{3}$, $\frac{7\pi }{6}$, and $\frac{4\pi }{3}$.
We know that the period of sine function is $2\pi$. So, in the interval there are two values at which the sine function is $\frac{\sqrt{3}}{2}$. One is $\frac{\pi }{3}$. And the sine is positive in quadrant II, thus the other value is: \begin{align} & \pi -\frac{\pi }{3}=\frac{3\pi -\pi }{3} \\ & =\frac{2\pi }{3} \end{align} Therefore, all the solutions to $\sin 2x=\frac{\sqrt{3}}{2}$ are given by: \begin{align} & 2x=\frac{\pi }{3}+2n\pi \\ & x=\frac{\pi }{6}+n\pi \end{align} Or, \begin{align} & 2x=\frac{2\pi }{3}+2n\pi \\ & x=\frac{\pi }{3}+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 }{6}+n\pi \\ & =\frac{\pi }{6}+0\times \pi \\ & =\frac{\pi }{6}+0 \\ & =\frac{\pi }{6} \end{align} \begin{align} & x=\frac{\pi }{6}+n\pi \\ & =\frac{\pi }{6}+1\times \pi \\ & =\frac{\pi +6\pi }{6} \\ & =\frac{7\pi }{6} \end{align} Or, 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 }{3}+n\pi \\ & =\frac{\pi }{3}+0\times \pi \\ & =\frac{\pi }{3}+0 \\ & =\frac{\pi }{3} \end{align} \begin{align} & x=\frac{\pi }{3}+n\pi \\ & =\frac{\pi }{3}+1\times \pi \\ & =\frac{\pi +3\pi }{3} \\ & =\frac{4\pi }{3} \end{align}