bad explanation in how to solve the cubic The question asked me to solve

$t^3+pt+q=0$ where $p,q$ real and $27q^2+4p^3<0$ using the identity $\cos{3\theta}=4\cos^3{\theta}-3\cos{\theta}$.

Answer goes like this and I have stopped where it just makes no sense at all to me.

Since the discriminant $-(27q^2+4p^3)>0$ implies there are three real roots this is the case here. Let $t=v \cos{\theta}$. Then
$$0=t^3+pt+q=v^3\cos^3{\theta}+pv\cos{\theta}+q \implies 4\cos^3{\theta}+\frac{4p}{v^2}\cos{\theta}+\frac{4q}{v^3}=0$$
as $p<0$ we may solve $v=\sqrt{-\frac{4p}{3}}$ for real $v$ and blah blah..

Whoa, where did that last statement randomly pop out from? What does it mean "as $p<0$"? Reverse engineering tells me that this only holds given that $1=-\frac{4p}{3v^2}$ is true. Where in heaven's name does that come from the equation above? Badly explained if you ask me, I cannot figure out how they Harry Potter-ed $v=\sqrt{-\frac{4p}{3}}$ right there.
What is going on? Does someone see the bizarre magic going on behind it? Great if someone would explain the trick to me because right now it just appeared out of thin air to me

Since $27q^2+4p^3<0$, $p^3<-(27/4)q^2<0$ so $p<0$.
