My previous question was about the cubic. Now, the following confuses me
Q. Use $\cos{3 \theta}=4 \cos^3{\theta}-3 \cos{\theta}$ to solve the cubic $t^3+pt+q=0$ for $p,q$ real when $27q^2+4p^3<0$
Here goes the solution
Since $27q^2+4p^3<0$ the cubic has three real roots. For this to be true, $-(27q^2+4p^3)>0$ and therefore $p<0$. If we let $t=u \cos{\theta}$ then our equation becomes
$$0=u^3 \cos^3{\theta}+pu\cos{\theta}+q \Rightarrow 0= 4 \cos^3{\theta}+\frac{4p}{u^2}\cos{\theta}+\frac{4q}{u^3}$$
since $p<0$,
Here comes what I don't get why
"we may solve $u=\sqrt{\frac{-4p}{3}}$ ($\leftarrow$ where did this come from?) for real $u$"
Reverse engineering tells me this is true if $\frac{4p}{u^2}+3=0$ is true. But what assures this? I thought of comparing coefficients.
So, $0=4 \cos^3{\theta}+\frac{4p}{u^2}\cos{\theta}+\frac{4q}{u^3}$ and $\cos{3 \theta}=4 \cos^3{\theta}-3 \cos{\theta}$ this gives, by rearranging since they both have $4\cos^3{\theta}$,
$$\cos{3 \theta}+3 \cos{\theta}=-\frac{4p}{u^2}\cos{\theta}-\frac{4q}{u^3}$$
Well, comparing coefficients doesn't seem right here because $\cos{3 \theta}$ gets in the way. Namely, sure I can collect the terms and get $(3+\frac{4p}{u^2})\cos{\theta}$ but nothing, nothing at all, assures me that this coefficient must be zero.
So I don't know where the solution made that leap in logic. It's really not apparent or obvious to me. Why is $3+\frac{4p}{u^2}=0$?
I'd do it in a slightly different way: set $t=u\cos\theta$; then the equation becomes $$ u^3\cos^3\theta+pu\cos\theta+q=0 $$ and we are still free to choose whatever value of $u\ne0$ we want. Dividing by $u^3$ and multiplying by $4$ we have $$ 4\cos^3\theta+\frac{4p}{u^2}\cos\theta+\frac{4q}{u^3}=0 $$ and we can choose $$ \frac{4p}{u^2}=-3 $$ that is, $u=\sqrt{-4p/3}$, and the equation becomes $$ 4\cos^3\theta-3\cos\theta=-\frac{4q}{u^3} $$ or $$ \cos3\theta=-\frac{4q}{u^3} $$ Note that $$ \left(-\frac{4q}{u^3}\right)^2=\frac{16q^2}{u^6}=\frac{16q^2}{(-4p/3)^3} \le1 $$ because of the hypothesis that the discriminant is negative. So the equation in the unknown $\theta$ has solutions.