Solve this trigonometric equation further

Question

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

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

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

Answer 1

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?