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?
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 $p$.