How to determine the order of the real roots of a cubic equation? This is a self-answered question (I didn't find a reference, and thought of documenting this). Consider the equation
$$
t^3+pt+q=0.
$$
Its discriminant is
$$
\Delta=-(4p^3+27q^2).
$$
Suppose that it has three distinct real roots; this is equivalent to $\Delta > 0$, or $4p^3+27q^2<0$.
In particular, this forces $p<0$.
These roots can be expressed as follows :
$$
x_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p} \sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right),
$$
where $k=0,1,2$.

Claim:
$$
x_0 > x_1 > x_2.
$$
How to prove this claim?

Comment:
In order to apply $\arccos$ we must have
$$-1 \le\frac{3q}{2p} \sqrt{\frac{-3}{p}} \le 1,$$
which is equivalent to
$$
\frac{27q^2}{-4p^3} \le 1.
$$
Since $p<0$, we can multiply this inequality by $-4p^3$, so it's equivalent to
$$
27q^2 \le -4p^3,
$$
or $\Delta \ge 0$.
 A: The three numbers $x_k=t\cos\left(\theta-k\frac{2\pi}3\right)$ ($k=0,1,2$) are supposed to be distinct and $t$ is positive.
Moreover, since $\theta\in[0,\pi/3],$ we have
$$\cos\theta-\cos\left(\theta-\frac{2\pi}3\right)=-2\sin\left(\theta-\frac\pi3\right)\sin\frac\pi3\ge0$$
and
$$\cos\left(\theta-\frac{2\pi}3\right)-\cos\left(\theta-\frac{4\pi}3\right)=-2\sin(\theta-\pi)\sin\frac\pi3\ge0.$$
Therefore,
$$x_0>x_1>x_2.$$
A: Set
$$
m:=\frac{1}{3}\arccos\left(\frac{3q}{2p} \sqrt{\frac{-3}{p}}\right),
$$
we have $0 \le m \le \pi/3$. (since the range of $\arccos$ is $[0,\pi]$).
The analysis in the question shows that $\Delta >0$ if and only if $-1<\frac{3q}{2p} \sqrt{\frac{-3}{p}}<1$, which is equivalent to the strict inequality $0<m<\pi/3$.
Then, up to a multiplication by the positive constant of $2\sqrt{-\frac{p}{3}}$, we have
$$
x_k=\cos\left(m-k\frac{2\pi}{3}\right),
$$
or
$$
x_0=\cos(m), x_1=\cos(m-\frac{2\pi}{3}),x_2=\cos(m-\frac{4\pi}{3}).
$$
Thus, $x_0 > x_1 > x_2$ is equivalent to
$$
\cos(m) >\cos(m-\frac{2\pi}{3})>\cos(m-\frac{4\pi}{3}),
$$
for any $m \in (0,\pi/3)$.
This can by differentiating the inequality $\cos(x) >\cos(x-\frac{2\pi}{3})$ and observing that the difference is monotone on $(0,\pi/3)$, and similarly for the second inequality.
