Trigonometric Identity Correction: Solving a Complex Equation

[anemone]

If $\dfrac{\sin 4x}{a}=\dfrac{\sin 3x}{b}=\dfrac{\sin 2x}{c}=\dfrac{\sin x}{d}$, show that $2d^3(2c^3-a^2)=c^4(3d-b)$.

 

[anemone]

If $\dfrac{\sin 4x}{a}=\dfrac{\sin 3x}{b}=\dfrac{\sin 2x}{c}=\dfrac{\sin x}{d}$, show that $2d^3(2c^3-a^2)=c^4(3d-b)$.

I'm sorry, folks!(Tmi)(Worried) But there's a misprint in the identity because it should read as $d^3(4c^2-a^2)=c^4(3d-b)$.

I must thank Opalg for letting me know something doesn't look right with the initial identity!(Sun)

 

[Saitama]

I'm sorry, folks!(Tmi)(Worried) But there's a misprint in the identity because it should read as $d^3(4c^2-a^2)=c^4(3d-b)$.

Spoiler

Let
$$\dfrac{\sin 4x}{a}=\dfrac{\sin 3x}{b}=\dfrac{\sin 2x}{c}=\dfrac{\sin x}{d}=\lambda$$

$$\sin 3x=3\sin x-4\sin^3 x \Rightarrow b\lambda =3d\lambda-4d^3\lambda^3 \Rightarrow \lambda^2=\frac{3d-b}{4d^3}$$

Also,
$$\sin 4x=2\sin 2x \cos 2x \Rightarrow 2c\lambda \sqrt{1-c^2\lambda^2}=a\lambda \Rightarrow \lambda^2=\frac{4c^2-a^2}{4c^4}$$

Equating the two expression for $\lambda^2$,
$$\frac{3d-b}{4d^3}=\frac{4c^2-a^2}{4c^4}$$
$$\Rightarrow d^3(4c^2-a^2)=c^4(3d-b)$$

$\blacksquare$