<s>[INST] $\sin(4x) = -2\sin(2x)$ 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.
 [/INST] 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.
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