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.