Common factor in $2\sin(x)\cos(x) + \sin(x) = 0$ I am stuck on part of a question :
The 1st line of work is :
$$2\sin(x)\cos(x) +\sin(x) = 0$$
The next line is :
$$\sin(x) \cdot (\sin(x)\cos(x) +1)$$
I see that $$\sin(x) \cdot 1 $$ gives $$\sin(x)$$
However how can $$\sin(x)\cdot \sin(x)\cos(x)$$ create $$2\sin(x)\cos(x)$$ 
As there are $2$ lots of $\sin(x)$ , however still only $1$ lot of $\cos(x)$?
Any help would be greatly appreciated.
This is the entire question :
 A: How can it? It cannot!
$$2\sin(x)\cos(x) + \sin(x) = 0$$ means
$$\sin(x)\left(2\cos(x) + 1\right) = 0$$
So you have the following cases (which are not coupled! They are two distinct possibilities)
$$
\begin{cases}
\sin(x)  = 0 \\\\
\cos(x)  = -\frac{1}{2}
\end{cases}
$$
Can you continue?
Remarks, just trig pills


*

*$\sin^2(x) + \cos^2(x) = 1$ for all $x\in\mathbb{R}$

*$\sin(2x) = 2\sin(x)\cos(x)$

*$\cos(2x) = \cos^2(x) - \sin^2(x)$


more here: https://en.wikipedia.org/wiki/List_of_trigonometric_identities
And now the solutions! (Spoiler alert)
$$\sin(x) = 0 \to x = n\pi ~~~~~~~ n\in\mathbb{Z}$$
$$\cos(x) = -\frac{1}{2} \to x = \frac{2}{3}(3\pi n \pm \pi) ~~~~~~~ n\in\mathbb{Z}$$
A: From $2\sin x\cos x+\sin x=0$ you get
$$
\sin x(2\cos x+1)=0
$$
because it's just like
$$
2ab+a=0
$$
that can be rewritten
$$
a(2b+1)=0
$$
By the way, there's a slicker way to solve $\sin2x+\sin x=0$. Rewrite it as
$$
\sin2x=-\sin x
$$
and recall that $-\sin x=\sin(x+\pi)$; therefore you have
$$
\sin 2x=\sin(x+\pi)
$$
that gives either
$$
2x=x+\pi+2k\pi \tag{1}
$$
or
$$
2x=\pi-(x+\pi)+2k\pi \tag{2}
$$
For the first solution set we get
$$
x=\pi+2k\pi \tag{1}
$$
and from the second one
$$
3x=2k\pi
$$
that is
$$
x=\frac{2}{3}k\pi \tag{2}
$$
If you want the solutions $-\pi\le x\le\pi$, you get, with $k=-1$ and $k=0$ in 1 and with $k=-1$, $k=0$ and $k=1$ from 2
$$
-\pi,\quad
-\frac{2}{3}\pi,\quad
0,\quad
\frac{2}{3}\pi,\quad
\pi
$$
