2

I thought this shouldn't be too hard, but evidently not. I am asked to solve $t^3+pt+q=0$ given $27q^2+4p^3<0$, using $\cos{3 \theta}=4 \cos ^3{\theta} - 3 \cos {\theta}$.

At first I thought, "Cardano's formula?" but this only comes to a dead end really, given its complexity, so I discarded the idea.

I then tried substitution, basically so that I can get some expression for $\theta$. So say, $t= -p \sqrt{p}\cos{\theta}$ or something similar. I only realized that I must let $t = i \sqrt{\frac{4p}{3}} \cos {\theta}$ in order to substitute in the original cubic equation so that I am left with an expression with $\cos{3 \theta}$ as the only unknown, and let $\Phi = 3 \theta$ and ultimately solve for $\theta$ and then for $t$.

However, the problem is that $i$ is not in the domain or range of $\cos$ as far as I know and thus no such $\Phi$ exists and therefore, my substitution is pointless.

I am unsure what else would work on this and out of ideas. How can this be soled at all with this trig identity?

John Trail
  • 3,279
  • why does cardano's formula not help? Btw i think this is meant to be solved using Vieta's approach to the cubic equation. – MrYouMath Jan 17 '16 at 16:46
  • I have not attempted to reproduce your calculations, but I'll point out that the cosine function does indeed extend to an analytic function on the entire complex plane. Perhaps your approach is not as pointless as you think? – Lee Mosher Jan 17 '16 at 17:36

1 Answers1

0

Hint: try $t=u\cos\theta$, so that you have $$u^3\cos^3\theta+pu\cos\theta+q=0$$ and then choose an appropriate value for $u$ based on $p$.

Théophile
  • 24,627
  • But to make $\cos {\theta}$ term disappear, $u$ will involve $i$. Is there any other way of interepreting the term "appropriate" .....?? – John Trail Jan 17 '16 at 16:59
  • Let $u = kp$, and find the value of $k$ that lets you use the trigonometric identity. A key observation is that we know $p < 0$ since $27q^2+4p^3<0$. This should ensure that $u$ remain real. – Théophile Jan 17 '16 at 17:08
  • right...then I get $k=\sqrt{\frac{4}{3 |p|}}$ then, as the negative sign should cancel out...is this correct? – John Trail Jan 17 '16 at 17:36
  • That's right. I would write this as $k = \sqrt\frac4{-3p}$; in your case, since $p < 0$, it amounts to the same thing, but $\sqrt\frac4{-3p}$ also works for the more general case, whereas $\sqrt\frac4{3|p|}$ would be wrong for $p>0$. – Théophile Jan 17 '16 at 17:47