Answer:
[tex]\displaystyle x = \frac{\pi}{3} +k\, \pi[/tex] or [tex]\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi[/tex], where [tex]k[/tex] is an integer.
There are three such angles between [tex]0[/tex] and [tex]2\pi[/tex]: [tex]\displaystyle \frac{\pi}{3}[/tex], [tex]\displaystyle \frac{2\, \pi}{3}[/tex], and [tex]\displaystyle \frac{4\,\pi}{3}[/tex].
Step-by-step explanation:
By the double angle identity of sines:
[tex]\sin(2\, x) = 2\, \sin x \cdot \cos x[/tex].
Rewrite the original equation with this identity:
[tex]2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0[/tex].
Note, that [tex]2\, (2\, \sin x \cdot \cos x)[/tex] and [tex](-2\, \sin x)[/tex] share the common factor [tex](2\, \sin x)[/tex]. On the other hand, [tex]2\sqrt{3}\, \cos x[/tex] and [tex](-\sqrt{3})[/tex] share the common factor [tex]\sqrt[3}[/tex]. Combine these terms pairwise using the two common factors:
[tex](2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0[/tex].
Note the new common factor [tex](2\, \cos x - 1)[/tex]. Therefore:
[tex]\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0[/tex].
This equation holds as long as either [tex]\left(2\, \sin x + \sqrt{3}\right)[/tex] or [tex](2\, \cos x - 1)[/tex] is zero. Let [tex]k[/tex] be an integer. Accordingly:
Any [tex]x[/tex] that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:
