## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $0$, $\frac{\pi }{3}$, $\pi$ and $\frac{4\pi }{3}$.
The period of the sine function is $2\pi$. In the interval there are two values at which the sine function is $\frac{1}{2}$. One is $\frac{\pi }{6}$ . The sine is positive in quadrant II, thus the other value is: \begin{align} & \pi -\frac{\pi }{6}=\frac{6\pi -\pi }{6} \\ & =\frac{5\pi }{6} \end{align} All the solutions to $\sin \left( 2x+\frac{\pi }{6} \right)=\frac{1}{2}$ are given by: \begin{align} & 2x+\frac{\pi }{6}=\frac{\pi }{6}+2n\pi \\ & 2x=2n\pi \\ & x=n\pi \end{align} Or, \begin{align} & 2x+\frac{\pi }{6}=\frac{5\pi }{6}+2n\pi \\ & 2x=\frac{4\pi }{6}+2n\pi \\ & x=\frac{2\pi }{6}+n\pi \\ & =\frac{\pi }{3}+n\pi \end{align} Where n is any integer. The solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$, $n=1$. The equation is calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows. \begin{align} & x=n\pi \\ & =0\times \pi \\ & =0 \end{align} \begin{align} & x=n\pi \\ & =1\times \pi \\ & =\pi \end{align} The second equation is also calculated 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 \\ & =\frac{\pi +3\pi }{3} \end{align} $=\frac{4\pi }{3}$