$\sin(4x) = -2\sin(2x)$ solutions in $[0,2\pi)$? My textbook gives the following answer:
$\sin(4x) + 2\sin(2x) = 0$

$2\sin(2x)\cos(2x) + 2\sin(2x) = 0$

$2\sin2x (\cos(2x) + 1) = 0$

$2\sin 2x = 0$

$\sin2x = 0$

$2x = πk$

$x = kπ/2$

In the interval $[0,2π)$ you have the solutions $0,π/2,π$ and $3π/2$. The book then shows the other solutions from $(\cos(2x) + 1) = 0.$
Here's my question: why is it $2x = πk$ and not $2x = 2πk$? 
I'm just in high school so I probably won't understand if there's a really complicated explanation.
 A: Your textbook is correct; however some remarks are in order.
First, setting $y=2x$, the equation becomes $\sin2y=2\cos y$; recalling the duplication formula, we get
$$
2\sin y\cos y=-2\cos y
$$
that can be reduced to
$$
\sin y(\cos y+1)=0
$$
Thus the equation splits into two:
$$
\sin y=0 \qquad\text{or}\qquad \cos y=-1
$$
However, if $\cos y=-1$, then $\sin y=0$, so all solutions of the second branch are solutions of the first branch.
Hence the only equation is $\sin y=0$, that is, $y=k\pi$, for $k$ an integer, which becomes
$$
2x=k\pi
$$
and so
$$
x=k\frac{\pi}{2}
$$
Thus the solutions in $[0,2\pi)$ are
$$
0,\quad
\frac{\pi}{2},\quad
\pi,\quad
\frac{3\pi}{2}
$$
In a different case you can't use such a “shorthand”: for instance, from $\sin y=\frac{1}{2}$ you get
$$
y=\frac{\pi}{6}+2k\pi
\qquad\text{or}\qquad
y=\pi-\frac{\pi}{6}+2k\pi
$$
However, in this special case $\sin y=0$ one has
$$
y=0+2k\pi
\qquad\text{or}\qquad
y=\pi-0+2k\pi=(2k+1)\pi
$$
so the first branch corresponds to even integer multiples of $\pi$ and the second branch to odd integer multiples of $\pi$ and we can abbreviate it in “$y$ is an integer multiple of $\pi$”, or $y=k\pi$.
You wouldn't be wrong if you use the “longer” path (with the branching).
A: Use the following identities
$$\sin(2 x) \equiv 2 \sin(x) \cos(x) $$
$$\cos(2 x) \equiv 2 \cos^2(x)-1 $$
$$\sin(4 x) \equiv \sin(2 x) \cos(2 x) = 8 \sin(x) \cos^3(x) - 4 \sin(x)\cos(x) $$
$$\sin(4 x) +2 \sin(2 x) = 8 \sin(x) \cos^3(x) - 4 \sin(x)\cos(x) + 4 \sin(x) \cos(x) $$
which simplifies to
$$ 8 \sin(x) \cos^3(x) =0 $$
with solution $x = k \frac{\pi}{2}$ since either $\sin(x)$ has to be zero or $\cos(x)$ has to be zero.
