Answer
The solution is $\underline{x=0,\frac{2\pi }{3},\pi ,\frac{4\pi }{3}}.$
Work Step by Step
We have to solve $\sin 2x+\sin x=0$:
$\begin{align}
& \sin 2x+\sin x=0 \\
& 2\sin x\cos x+\sin x=0
\end{align}$
Factor:
$\sin x\left( 2\cos x+1 \right)=0$
So, from the above result, it can be concluded that,
$\sin x=0$
And the other value will be:
$2\cos x+1=0$
With the general formula:
$\begin{align}
& \sin x=0 \\
& x=n\pi
\end{align}$
Compute the value of $\cos x$.
$\begin{align}
& 2\cos x+1=0 \\
& 2\cos x=-1 \\
& \cos x=\frac{-1}{2}
\end{align}$
So, in the interval $\left[ 0,2\pi \right),$ the cosine function is $\frac{-1}{2}$ at $\frac{2\pi }{3}\text{ and }\frac{4\pi }{3}$ according to the trigonometric table.
$\begin{align}
& \cos x=\frac{-1}{2} \\
& x=\frac{2\pi }{3}
\end{align}$
The other value is:
$\begin{align}
& \cos x=\frac{-1}{2} \\
& x=\frac{4\pi }{3}
\end{align}$
Since the period of the cosine function is $2\pi $, the general solution of the equation is:
$x=\frac{2\pi }{3}+2n\pi \,\text{ and }\frac{4\pi }{3}+2n\pi $
To get different solutions, $\text{put }n=0,1,2,3\ldots $
$\begin{align}
& \text{For }n=0, \\
& x=n\pi \\
& =0
\end{align}$
So, another value is:
$\begin{align}
& x=\frac{2\pi }{3}+2n\pi \\
& =\frac{2\pi }{3}
\end{align}$
Other value of x is:
$\begin{align}
& x=\frac{4\pi }{3}+2n\pi \\
& =\frac{4\pi }{3}
\end{align}$
Substitute the value of n:
$\begin{align}
& \text{For }n=1, \\
& x=n\pi \\
& =\pi
\end{align}$
When putting the other values of $n$, the solution becomes outside the interval $\left[ 0,2\pi \right).$
Thus, the exact solution of the equation $\sin 2x+\sin x=0$ on the interval $\left[ 0,2\pi \right)$ is $\underline{x=0,\frac{2\pi }{3},\pi ,\frac{4\pi }{3}}.$
