**Solution**.

With the substitution $t = \frac15 + s$, the equation is written as
$$20s^3 - 18s^2 + qs - r = 0, \tag{1}$$
where
\begin{align*}
	q &:= \frac{12}{5} - 4 \cos^2 \alpha - 4 \cos^2 \beta - 4 \cos^2 \gamma, \\
	r &:= - \frac{9}{25} - \frac15\cos^2 \alpha - \frac15 \cos^2 \beta - \frac15 \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma.
\end{align*}

We claim that the cubic equation (1) has at least one real root $s \le 0$. 
Assume, for the sake of contradiction, that (1) has three positive real roots. Then we have $q > 0, r > 0$.
However, we can prove that 
$\frac54 r + \frac{3}{16}q \le 0$ that is
$$- \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + \frac52 \cos \alpha \cos \beta \cos \gamma \le 0. \tag{2}$$
(2) is true since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \ge 3\sqrt[3]{\cos^2 \alpha  \cos^2 \beta  \cos^2 \gamma} \ge 3|\cos \alpha \cos \beta \cos \gamma| \ge \frac52 \cos \alpha \cos \beta \cos \gamma$. Contradiction. Thus, the claim is proved.

We are done.