solve ;$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$ This is one more of my unsolved trigonometry  questions:

solve ;$$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$$


My Try
provided that $\cos (2x-2\pi/3)\neq 0$ and squaring both sides $$1-4\cos^2 4x=8\cos (2x-2\pi/3)\cos^2(2x-\pi/6)....(1)$$
$$1-4\cos^2 4x=4\cos(2x-2\pi/3)(1+\cos(4x-\pi/3))...(2)$$
For convenience we take $2x=t$:$$1-4\cos^2 2t=4\cos(t-2\pi/3)(1+\cos(2t-\pi/3))....(3)$$
$$=4\cos(t-2\pi/3)-2\cos(3t)+2\cos (t+\pi/3)....(4)$$
What do i do next?
Source:A Panchishkin- Trigonometric functions

It's been a tour de force, mate. Even with the help of Wolfram Mathematica.
After expanding with standard formulae and subbed $2x=u$ I got this mess
$$-2 \sin ^4 u-\sqrt{3} \sin  u-2 \cos ^4 u+2 \cos ^3 u+\cos  u+12 \sin ^2 u \cos ^2 u-6 \sin ^2 u \cos  u-1=0$$
I substituted $$\sin u=\frac{2t}{1+t^2};\;\cos u=\frac{1-t^2}{1+t^2};\;t=\tan u/2$$
After a while I got
$$t \left(3 t^7+\sqrt{3} t^6-39 t^5+3 \sqrt{3} t^4+73 t^3+3 \sqrt{3} t^2-13 t+\sqrt{3}\right)=0$$
And thanks to Mathematica this amazing factorization, using the function
Factor[t (Sqrt[3] - 13 t + 3 Sqrt[3] t^2 + 73 t^3 + 3 Sqrt[3] t^4 - 
    39 t^5 + Sqrt[3] t^6 + 3 t^7), Extension -> Automatic]

$$\left(\sqrt{3}-t\right) t \left(t+\sqrt{3}-2\right) \left(t+\sqrt{3}+2\right) \left(3 t+\sqrt{3}\right) \left(-3 t^3+3 \sqrt{3} t^2+9 t-\sqrt{3}\right)=0$$
The linear factors are trivial. The cubic is another mess. I got these three approximate solutions
$t_1=-1.19175, t_2=0.176327, t_3=2.74748$
A little consolation is that $t=\tan x$
In $[0,2\pi]$ solutions are $$x=0,\frac{5 \pi }{6},\pi ,\frac{4 \pi }{3},\frac{11 \pi }{6},2 \pi,0.174533,2.26893,5.41052,1.22173$$
