<s>[INST] $\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{\sin{4x}}{4}$ - solving a trigonometric equation The question is to solve the following equation:
$$ \sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{\sin{4x}}{4} $$ 
There is a tedious and mistake-prone way to do this, that is using trigonometric identities to write the equation in terms of $\sin{x}$ exclusively. But what I'm concerned about, is if there's another, perhaps somewhat clever way to deal with such problems?
 [/INST] $$\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{2\cdot\sin{2x}\cdot \cos 2x}{4}$$
$$2\cdot\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\sin{2x}\cdot \cos 2x $$
$$\sin 2x(2\cdot\sin{x}\cdot\sin{3x}-\cos 2x) =0$$
$$\sin 2x(-\cos 4x+\cos 2x-\cos 2x)=0 \Rightarrow \sin2x \cdot\cos4x=0$$
$$\sin 2x =0 \Rightarrow x=\frac{k\pi}{2}$$
$$\cos 4x =0 \Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}$$
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