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Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let $$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$ $$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$ Then $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2} \\ &=:g(u)\ge g(3)=0 \end{aligned}$$ (the latter inequality holds because $u=3$ is the only critical point of $g$ and $g(0)>0$),
$$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\ &=:h(u)\ge\min(h(0),h(3/5)) \\ &>0 \text{ if }0\le u<3/5 \end{aligned}$$ (the 2nd inequality in the latter display holds because $h$ is concave). So, $p(\min(t_*,1/5))\ge0$.

Also, $p(-\infty+)=-\infty<0$.

Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$

Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let $$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$ $$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$ Then $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2} \\ &=:g(u)\ge g(3)=0 \end{aligned}$$ (the latter inequality holds because $u=3$ is the only critical point of $g$ and $g(0)>0$),
$$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\ &=:h(u)\ge\min(h(0),h(3/5)) \\ &>0 \text{ if }0\le u<3/5 \end{aligned}$$ (the 2nd inequality in the latter display holds because $h$ is concave). So, $p(\min(t_*,1/5))\ge0$.

Also, $p(-\infty+)=-\infty<0$.

Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$

Note that no conditions on $a,b,c$ (that is, on $\alpha,\beta,\gamma$) were needed or used here.

Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let $$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$ $$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$ Then $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2} \\ &=:g(u)\ge g(3)=0 \end{aligned}$$ (the latter inequality holds because $u=3$ is the only critical point of $g$ and $g(0)>0$),
$$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\ &=:h(u)\ge\min(h(0),h(3/5)) \\ &>0 \text{ if }0\le u<3/5 \end{aligned}$$ (the 2nd inequality in the latter display holds because $h$ is concave), Also, $p(-\infty+)=-\infty<0$.

Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$

Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let $$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$ $$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$ Then $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2} \\ &=:g(u)\ge g(3)=0 \end{aligned}$$ (the latter inequality holds because $u=3$ is the only critical point of $g$ and $g(0)>0$),
$$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\ &=:h(u)\ge\min(h(0),h(3/5)) \\ &>0 \text{ if }0\le u<3/5 \end{aligned}$$ (the 2nd inequality in the latter display holds because $h$ is concave). So, $p(\min(t_*,1/5))\ge0$.

Also, $p(-\infty+)=-\infty<0$.

Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$

Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let $$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$ $$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$ Then $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2}=:g(u)\ge g(3)=0, \end{aligned}$$ $$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\ &=:h(u)\ge\max(h(0),h(3/5))>0 \text{ if }0\le u<3/5. \end{aligned}$$ Also, $p(-\infty+)=-\infty<0$.

Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$

Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let $$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$ $$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$ Then $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2} \\ &=:g(u)\ge g(3)=0 \end{aligned}$$ (the latter inequality holds because $u=3$ is the only critical point of $g$ and $g(0)>0$),
$$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\ &=:h(u)\ge\min(h(0),h(3/5)) \\ &>0 \text{ if }0\le u<3/5 \end{aligned}$$ (the 2nd inequality in the latter display holds because $h$ is concave), Also, $p(-\infty+)=-\infty<0$.

Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$