## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{6},\,\frac{5\pi }{6},\,\frac{7\pi }{6}\,,\text{ and }\ \frac{11\pi }{6}$.
We know that when the "mod" is removed from the equation, it is needed to consider both the positive and the negative values. So, in order to solve the equation on the interval $[0,2\pi )$ , the following will be the course of action: \begin{align} & \left| \text{sin}\ x \right|=\frac{1}{2} \\ & \text{sin}\ x=\frac{1}{2} \end{align} or $\sin x=-\frac{1}{2}$ Then, solve for $x$ on the interval $[0,2\pi )$. So, in the quadrant graph, the value of sine is $\frac{1}{2}$ at $\frac{\pi }{6}$ and $\frac{5\pi }{6}$. This implies, \begin{align} & sinx=sin\frac{\pi }{6} \\ & x=\frac{\pi }{6} \end{align} \begin{align} & sinx=sin\frac{5\pi }{6} \\ & x=\frac{5\pi }{6} \end{align} So, in the quadrant graph, the value of sine is $-\frac{1}{2}$ at $\frac{7\pi }{6}$ and $\frac{11\pi }{6}$. This implies, \begin{align} & \sin x=sin\frac{7\pi }{6} \\ & x=\frac{7\pi }{6} \end{align} \begin{align} & \sin x=sin\frac{11\pi }{6} \\ & x=\frac{11\pi }{6} \end{align} These are the proposed solutions of the cosine functions. Thus, the actual solutions in the interval $[0,2\pi )$ will be $\frac{\pi }{6},\,\frac{5\pi }{6},\,\frac{7\pi }{6},\,\text{ and }\ \frac{11\pi }{6}$.