The equation $$ \sin x + \sqrt{3} \cos x = 1,$$ where $x \in [-2\pi, 2\pi]$ has four roots $-\dfrac{3\pi}{2}$, $\dfrac{\pi}{2}$, $\dfrac{\pi}{2}$ and $\dfrac{11\pi}{6}$. Sum of roots of the given equation is $\dfrac{2\pi}{3}$. How do I tell Mathematica to do that?
I tried
Solve[{Sin[x] + Sqrt[3]*Cos[x] == 1, -2*\[Pi] <= x <= 2*\[Pi]}, {x}]
edit from comment
I repaired the given equation to
Solve[{Sin[x] + Sqrt[3]Cos[x] == 1, -2[Pi] <= x <= 2*[Pi]}, {x}]
and used
Simplify@Total[ x /. Solve[{Sin[x] + Sqrt[3] Cos[x] == 2, -2 Pi <= x <= 2 Pi}, {x}]],
and I got the answer
-4 [Pi] + 8 ArcTan[1/(2 + Sqrt[3])].
How do I get the answer -((5 [Pi])/3)?