Using $A+B+C=\pi$, we have
$\begin{align*}\sin 2A+\sin 2B+\sin 2C&=\sin 2A+\sin 2B-\sin (2A+2B)\\&=2\sin (A+B)\cos (A-B)-2\sin(A+B)\cos(A+B)\\&=2\sin(A+B)[\cos(A-B)-\cos(A+B)]\\&=2\sin(A+B)(-2\sin A \sin B)\\&=2\sin C (2\sin A\sin B)\\&=4\sin A \sin B\sin C\end{align*}$
Replace $A,\,B$ and $C$ in the above calculation by $2A,\,2B$ and $2C$ to get
$\sin 4A+\sin 4B+\sin 4C=-4\sin 2A\sin 2B\sin 2C$
Using these two inequalities, the given relation is equivalent to
$-4cos A\cos B\cos C=1$
Since the product of cosines in $-4\cos A\cos B\cos C=1$ is negative, and $A\ge B\ge C$, we must have $A>\dfrac{\pi}{2}>2C>0$.
Using $A+B+C=\pi$, from $-4\cos A\cos B\cos C=1$, we get
$4\cos A \cos(A+C)\cos C=1$ and so
$4\cos A(\cos A \cos C-\sin A \sin C)\cos C=1$
Divide both sides by $\cos^2 C$ to get
$4\cos^2 A-4\sin A\cos A\tan C=\dfrac{1}{\cos^2 C}$
Let $x=\cos A$ and $t=\tan^2 C$.
From $A>\dfrac{\pi}{2}>2C>0$, it follows that $-1<x<0$ and $0<t<1$.
Rewrite $4\cos^2 A-4\sin A\cos A\tan C=\dfrac{1}{\cos^2 C}$ using this notation and rearrage to get
$4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$
Square both dies and move all the terms to one side to get
$16(1+t)x^4-8(1+3t)x^2+(t+1)^2=0$
Applying the quadratic formula, we get
$x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$
Since $-1<x<0$ and $0<t<1$, $4x^2-(1+t)<0$ whenever $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ holds.
For the solutions obtained from the quadratic formula we have
$4x^2-(1+t)=\dfrac{(t-t^2)\pm \sqrt{3t+6t^2-t^3}}{1+t}$ and since $t-t^2>0$ it is clear that when $x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$, $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ is not satisfied.
Therefore, since $x<0$, we must have
$x=-\dfrac{1}{2}\sqrt{\dfrac{1+3t-\sqrt{3t+6t^2-t^3}}{1+t}}$
We want to find the range of values of $x$ for $t\in(0,\,1)$. Implicitly differentiate $x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$ to get
$x\dfrac{dx}{dt}=\dfrac{4\sqrt{3t+6t^2-t^3}+t^3+3t^2-9t-3}{16(1+t)^2\sqrt{3t+6t^2-t^3}}$
There is no $t\in(0,\,1)$ for which either $x=0$ or $16(1+t)^2\sqrt{3t+6t^2-t^3}=0$, so we conclude that the critical points of $x$ satisfy
$4\sqrt{3t+6t^2-t^3}=-t^3-3t^2+9t+3\\(3+9t-3t^2-t^3)^2-16(3t+6t^2-t^3)=0\\t^6+6t^5-9t^4-44t^3-33t^2+6t+9=0\\(t-3)(t+1)^3(t^2+6t-30$
The only critical point in the range $[0,\,1]$ is $t=2\sqrt{3}-2$. The corresponding value of $x$, obtained after a tedious but straightforward calculation is $\dfrac{1=\sqrt{3}}{2}$. From
$x=-\dfrac{1}{2}\sqrt{\dfrac{1+3t-\sqrt{3t+6t^2-t^3}}{1+t}}$,
we easily evaluate
$\displaystyle \lim_{t\rightarrow 0^+} x=-\dfrac{1}{2}$ and
$\displaystyle \lim_{t\rightarrow 1^-} x=-\dfrac{\sqrt{2-\sqrt{2}}}{2}$
allowing us to conclude that for $t\in(0,\,1)$, we have
$-\dfrac{1}{2}<x\le \dfrac{1-\sqrt{3}}{2}$
Finally, we check that for each $x$ in this interval there is a corresponding triangle whose angles $A,\,B$ and $C$ satisfy the given relation. Suppose $x_0$ is such that $\dfrac{1}{2}<x_0\le \dfrac{1-\sqrt{3}}{2}$. Let $A=\cos ^{-1}x_0$. Since $\cos^{-1}$ is a decreasing function we have $A<\cos^{-1}(-0.5)=\dfrac{2\pi}{3}$.
By the intermediate value theorem, since $x$ is continuous on $0,\,2\sqrt{3}-3]$, there exists a $t+0$ in this interval such that $x_0=x(t_0)$.
Let $C=\tan^{-1}\sqrt{t_0}$. Note that $\sqrt{t_0}\le \sqrt{2\sqrt{3}-3}<\sqrt{3}$, where $C<\dfrac{\pi}{3}$.
Let $B=\pi-A-C$, the earlier comments about the ranges for $A$ and $C$ imply $B>0$.
We claim that a triangle with angles $A,\,B$ and $C$ satisfies the relation given in the problem. From the construction, $\cos C=(1+\tan^2 C)^{-0.5}=(1+t_0)^{-0.5}$ and $\cos A=x_0$. Moreover, $x_0$ and $t$ satisfy equation $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$. We calculate, using trigonometric equalities to evaluate $\sin A$ and $\sin C$.
$\begin{align*}\cos B&=\cos (\pi-(A+C))\\&=-\cos (A+C)\\&=-\cos A\cos C+\sin A \sin C\\&=-x_0\sqrt{\dfrac{1}{1+t_0}}+\sqrt{1-x_0^2}\cdot \sqrt{1-\dfrac{1}{1+t_0}}\\&=\sqrt{\dfrac{1}{1+t_0}}(-x_0+\sqrt{1-x_0^2}\cdot \sqrt{t_0})\end{align*}$
Since $x_0\ne 0$, we rearrange $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ to get $\sqrt{1-x_0^2}\cdot \sqrt{t_0}=x_0-\dfrac{1+t_0}{4x_0}$.
Hence $\cos B=\sqrt{\dfrac{1}{1+t_0}}\cdot \dfrac{1+t_0}{-4x_0}=\dfrac{\cos C(1+t_0)}{-4\cos A}$.
It follows that $-4\cos A\cos B\cos C=1$, so $A,\,B,\,C$ satisfy $-4cos A\cos B\cos C=1$, which is equivalent to the equality in the question.
Therefore, we conclude that the possible range of value for $\cos A$ is given by $-\dfrac{1}{2}<\cos A\le \dfrac{1-\sqrt{3}}{2}$.
