How to solve this trigonometric equation further? Find the sum of the roots of the equation
$$
\sin (2\cos x - \sqrt{3}) = 0
$$
belonging to the segment
$$
A_m = [2m\pi; ~ \frac{\pi}{2} + (2m+2)\pi]
$$
where $m = -4$. Round the number to two decimal places if necessary.
my incomplete solution:

$$2\cos x - \sqrt 3 = k\pi$$
$$k = 0 \text{ fits}$$
$$k = 1$$
$$ 2\cos x = pi + \sqrt3 = 3.14 + 1.73> 2$$ - not suitable. For
larger $k$, it is even worse ...
$$k = -1$$
$$2\cos x = -3.14 + 1.73 = -1.41$$ Hmm Suitable unfortunately.
Smaller ones won't fit
This means that there are 2 options

*

*$$cos x = 0$$ $$x = \pi / 2 + 2n\pi$$ The segment $$[-8\pi, \pi / 2 - 6\pi]$$ 2
roots hit $$-8\pi + \pi / 2$$ and $$-6\pi + \pi / 2$$


*$$\cos x = -0.705$$ (coincidence with the root of 2 is random here) But
approximately you can say $$x = + -2\pi / 3 + 2n\pi$$

I have a problem with subsequent calculations, it lies in the fact that I do not understand how and what to calculate
I would be grateful if you could solve this problem to the end with an explanation
 A: Hint:
For real $x,$
$$-2-\sqrt3\le2\cos x-\sqrt3\le2-\sqrt3=\dfrac1{2+\sqrt3}$$
$\dfrac1{2+\sqrt3}<\pi\implies k<1$
Again, $2+\sqrt3<2\pi\implies k>-2$
So, what are the possible values of $k?$
Can you take it from here?
A: $$2\cos x-\sqrt3=k\pi$$ is possible with $k=-1$ and $k=0$.
Now we have the solutions
$$x=2n\pi\pm\arccos\frac{\sqrt3}2$$ and $$x=2n\pi\pm\arccos\frac{\sqrt3-\pi}2.$$
As solutions are requested in the range $\left[-8\pi,-8\pi+\dfrac52\pi\right]$ you need to find suitable values of $n$. Start with $n=-4$ (which gives two solutions) and increase $n$ until the options become too large.

A: based on the data given by the user Yves Daoust, I solved this problem

5 roots fall into the specified interval from the graph.
the interval [-8pi, pi / 2 - 6pi] is divided into two: [-8pi, - 6pi]
and [-6pi, - 6pi + pi / 2].
The first interval will have 4 roots:
$$x = pi / 6 - 8pi$$
$$x = -pi / 6 - 6pi$$
$$x = -7pi + arccos ((pi-sqrt3) / 2)$$
$$x = -7pi - arccos ((pi-sqrt3) / 2)$$
The second interval [-6pi, - 6pi + pi / 2] corresponds on the unit
circle to the first coordinate quarter, so $$x = pi / 6 - 6pi$$ will get
here.
The sum of these five roots is $$pi (-8-6-7-7-6 + 1/6) = - pi * 203/6 =
> -106, 29.$$

to check the result

