## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{2}$, $\frac{7\pi }{6}$, and $\frac{11\pi }{6}$.
We have to solve the equation on the interval $[0,2\pi )$; the following will be the course of action. Start with: $2{{\sin }^{2}}x-\sin x-1=0$ Solve it as: \begin{align} & 2{{\sin }^{2}}x-2\sin x+\sin x-1=0 \\ & 2\sin x\left( \sin x-1 \right)+1\left( \sin x-1 \right)=0 \\ & \left( 2\sin x+1 \right)\left( \sin x-1 \right)=0 \end{align} Each factor needs to be calculated as: \begin{align} & 2\sin x+1=0 \\ & 2\sin x=0-1 \\ & \sin x=-\frac{1}{2} \end{align} Or, \begin{align} & \sin x-1=0 \\ & \sin x=0+1 \\ & \sin x=1 \end{align} Then, solve for $x$ on the interval $[0,2\pi )$. In the quadrant graph, the value of sine is $1$ at the angle of $\frac{\pi }{2}$. It implies, \begin{align} & \sin x=\sin \frac{\pi }{2} \\ & x=\frac{\pi }{2} \end{align} In the quadrant graph, the value of sine is $-\frac{1}{2}$ at the angle of $\frac{7\pi }{6}$, and $\frac{11\pi }{6}$. It 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}