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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?

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$$\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\frac{2\cdot\sin{2x}\cdot \cos 2x}{4}$$

$$2\cdot\sin{x}\cdot\sin{2x}\cdot\sin{3x}=\sin{2x}\cdot \cos 2x $$

$$\sin 2x(2\cdot\sin{x}\cdot\sin{3x}-\cos 2x) =0$$

$$\sin 2x(-\cos 4x+\cos 2x-\cos 2x)=0 \Rightarrow \sin2x \cdot\cos4x=0$$

$$\sin 2x =0 \Rightarrow x=\frac{k\pi}{2}$$

$$\cos 4x =0 \Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}$$

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Only using the double angle formula:

$$\frac{\sin4x}4=\frac{2\sin2x\cos2x}4$$

Thus, either $\sin2x=0$, or

$$2\sin x\sin3x=\cos2x$$

And then this last bit can be done by using the triple angle formula for $\sin$ and the other double angle formula.

$$8\sin^4x+6\sin^2x=1-2\sin^2x$$

Let $\sin^2x=u$

$$0=8u^2+8u-1,\ u>0\implies u=\frac{-2+\sqrt6}4$$

And lastly recall that $\cos2x=1-2u$.

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