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According to this post, we can solve a cubic equation

$$t^3+pt+q=0$$

by the trigonometric identity

$$4\cos^3\theta-3\cos\theta-\cos3\theta=0$$

So I've tried to solve the quartic equation

$$t^4+pt^2+qt+r=0$$

using the identity

$$8\cos^4\theta-4\cos2\theta-\cos4\theta-3=0$$

$$8\cos^4\theta-4(2\cos^2\theta-1)-\cos3\theta\cos\theta+\sin3\theta\sin\theta-3=0$$

$$8\cos^4\theta-8\cos^2\theta-\cos3\theta\cos\theta+1+\sin3\theta\sin\theta=0$$

If we let $t:=A\cos\theta$, then we have

$$A^4\cos^4\theta+A^2p\cos^2\theta+Aq\cos\theta+r=0$$ or $$8\cos^4\theta+\frac{8p}{A^2}\cos^2\theta+\frac{8q}{A^3}\cos\theta+\frac{8r}{A^4}=0$$

now $\frac{8p}{A^2}=-8$ implies that $A=\sqrt{-p}$ and we need to find $\theta$ in the following system

$$\begin{cases}\cos3\theta=-\frac{8q}{A^3}\\1+\sin3\theta\sin\theta=\frac{8r}{A^4}\end{cases}$$

What would you think on solving this system? Is it possible to solve this system? I couldn't find anything on the web about the solving a quartic equation by trigonometric power identity, so I don't know if this way works ...

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In general, you cannot parameterize a two-dimensional set, like $$ \left(-\frac{8q}{A^3},\frac{8r}{A^4}\right) $$ with a one-edimensional curve, like $$ (\cos3\theta,1+\sin3\theta\sin\theta) $$

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  • $\begingroup$ Would you say that this system doesn't have any answer? $\endgroup$ – user486600 Nov 18 '17 at 17:37

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