Answer
The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{2}$, $\frac{7\pi }{6}$, and $\frac{11\pi }{6}$.
Work Step by Step
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,
$\begin{align}
& \sin x=\sin \frac{7\pi }{6} \\
& x=\frac{7\pi }{6}
\end{align}$
$\begin{align}
& \sin x=\sin \frac{11\pi }{6} \\
& x=\frac{11\pi }{6}
\end{align}$
