This question is the same (or similar) as this one but in that question, the answer "use the trigonometric method" is not explained further in the Wikipedia page (it was 6 years ago so the page has probably been heavily edited since this question) so it needs to be asked again or at least edited up to date.

Take, for example, equation $x^3 - 7x + 5 = 0$

This is the graph of the cubic curve:

enter image description here

The equation clearly has three real roots. However, the equation is not factorable, so to solve it we would have to resort to a cubic formula. Let's use the mainstream cubic formula (not depressed) here: Eventually after substituting $a = 1$, $b = 0$, $c = -7$ and $d = 5$ into all three solutions we get (from WolframAlpha):

enter image description here

Even though all three solutions are real solutions, they all have complex closed forms (in their non-trigonometric states), a separate question I asked a few hours ago explained how to convert some specific closed forms into their trigonometric forms using Euler's formula (which I now understand) - but is this 'rule' applicable to all examples of real numbers with seemingly complex closed expressions? With enough time, could I theoretically remove the imaginary numbers from all three real solutions, and in their place substitute matching trigonometric identities?

  • 4
    $\begingroup$ See casus irreducibilus $\endgroup$
    – user170231
    Commented Feb 2, 2023 at 2:14
  • $\begingroup$ @user170231 That is incredible. $\endgroup$
    – user1112591
    Commented Feb 2, 2023 at 2:37
  • $\begingroup$ Look at my edit for the hyperbolic solution $\endgroup$ Commented Feb 2, 2023 at 3:42

2 Answers 2


The answer is Yes.

$$x^3 - 7x + 5 = 0$$

Now, follow the steps given here for $a=1$, $b=0$, $c=-7$ and $d=5$.

You have $$\Delta=697 \qquad \qquad p=7\qquad \qquad q=5$$

$\Delta >0$ means three real roots. So, apply the trigonometric solution to obtain $$x_k=2 \sqrt{\frac{7}{3}} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(-\frac{15 }{14}\sqrt{\frac{3}{7}}\right)\right)\right)\quad \text{for} \quad k=0,1,2$$

Isn't a bit nicer than what Wolfram Alpha gave ?


Let me do the same for $$x^3 + 7x + 5 = 0$$

Follow the steps for $a=1$, $b=0$, $c=7$ and $d=5$.

You have $$\Delta=-2047 \qquad \qquad p=7\qquad \qquad q=5$$

$\Delta <0$ means one real roots. So, apply the hyperbolic solution to obtain $$x_r=-2 \sqrt{\frac{7}{3}} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{15}{14} \sqrt{\frac{3}{7}}\right)\right)$$

  • $\begingroup$ Definitely, this is one of the most exciting mathematical properties I've come across, much more interesting than Euler's formula. $\endgroup$
    – user1112591
    Commented Feb 2, 2023 at 2:41

In general, for an $n$'th root of a complex number $a+bi$, write $a+bi$ in polar form as $r e^{i\theta}$ where $r = \sqrt{a^2+b^2}$, $\cos(\theta) = a/r$ and $\sin(\theta) = b/r$, then the $n$'th roots of $a+bi$ are $$r^{1/n} e^{i(\theta + 2\pi j)/n} = r^{1/n} (\cos((\theta + 2\pi j)/n) + i \sin((\theta + 2\pi j)/n)), \ j = 0, \ldots, n-1$$


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