1ª forma :

[tex]\displaystyle \sf sen(x)+\sqrt{3}\cdot cos(x) = 1 \\\\ 2\cdot \left(sen(x) \cdot \frac{1}{2}+\frac{\sqrt{3}}{2}\cdot cos(x) \right) = 1 \\\\\\ \underbrace{\sf sen(x)\cdot cos\left(\frac{\pi}{3}\right)+sen\left(\frac{\pi}{3}\right)\cdot cos(x)}_{\displaystyle \sf sen\left(x+\frac{\pi}{3}\right)} = \frac{1}{2} \\\\\\\\ sen\left(x+\frac{\pi}{3}\right) = sen\left(\frac{\pi}{6}\right)= sen\left(\frac{5\pi}{6}\right)[/tex]

[tex]\displaystyle \sf x+\frac{\pi}{3} = \frac{\pi}{6} \to x= \frac{\pi}{6}-\frac{\pi}{3} \to \boxed{\sf x =\frac{-\pi}{6} } \\\\\\ x+\frac{\pi}{3} = \frac{5\pi}{6} \to x=\frac{5\pi}{6}-\frac{\pi}{3} \to \boxed{\sf x= \frac{3\pi}{6}}[/tex]

Ante de fazer a soma, note que o intervalo é de [0,2π]. Então não podemos usar o ângulo negativo.
Sabemos que -π/6 está o 4º quadrante, para deixa-lo positivo basta irmos até o 2π e voltar π/6, ou seja :

[tex]\displaystyle \sf \frac{-\pi}{6} = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}[/tex]

Somando as soluções :
[tex]\displaystyle \sf S = \frac{11\pi}{6}+\frac{3\pi}{6} \\\\\\ S = \frac{14\pi}{6} \\\\\\ \Large\boxed{\sf \ S = \frac{7\pi}{3}\ }\checkmark[/tex]

2ª forma :

[tex]\displaystyle \sf sen(x) +\sqrt{3}\cdot cos(x) = 1 \ \ ; \ \ x \ \in \ [0,2\pi] \\\\ \sqrt{3}\cdot cos(x) = 1-sen(x) \\\\ \left(\sqrt{3}\cdot cos(x) \right)^2 = \left(1-sen(x) \right) ^2 \\\\ 3\cdot \underbrace{\sf cos^2(x)}_{1-sen^2(x)} = 1- 2\cdot sen(x) +sen^2(x) \\\\\\ 3\cdot (1-sen^2(x) ) = sen^2(x) -2\cdot sen(x) +1 \\\\ 3-3\cdot sen^2(x) = sen^2(x) -2\cdot sen(x) +1 \\\\ sen^2(x) +3\cdot sen^2(x) -2\cdot sen(x) +1 -3 = 0 \\\\ 4\cdot sen^2(x) -2\cdot sen(x) -2 = 0[/tex]

[tex]\displaystyle \sf 2\cdot sen^2(x) -sen(x)-1 = 0 \\\\ sen(x) = \frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 2\cdot(-1)}}{2\cdot 2} \\\\\\ sen(x) =\frac{1\pm\sqrt{1+8}}{4} =\frac{1\pm\sqrt{9}}{4} \\\\\\ sen(x) = \frac{1+3}{4} \to sen(x) = 1 \to\boxed{\sf x = \frac{\pi}{2} } \\\\\\ sen(x) = \frac{1-3}{4} \to sen(x) =\frac{-1}{2} \to \boxed{\sf x= \frac{7\pi}{6} \ ; \ x=\frac{11\pi}{6} } \\\\\\ \text{note que o \'unico valor que ao substituir na equa\c c\~a n\~ao se verifica \'e o } \frac{7\pi}{6} :[/tex]

[tex]\displaystyle \sf sen\left(\frac{7\pi}{6}\right) +\sqrt{3}\cdot cos\left(\frac{7\pi}{6}\right) = 1 \\\\\\ \frac{-1}{2} +\sqrt{3}\cdot \left(\frac{-\sqrt{3}}{2}\right) = 1 \\\\\\ \frac{-1-3}{2} = 1 \to (ABSURDO) \\\\\\ \text{Portanto as solu\c c\~oes s\~ao} : \\\\ x= \frac{\pi}{2} = \frac{3\pi}{6} \\\\\\ x= \frac{11\pi}{6} \\\\\\ somando-as : \\\\ S = \frac{11\pi}{6} + \frac{3\pi}{6} = \frac{14\pi}{6} = \frac{7\pi }{3} \\\\\\ \Large\boxed{\sf\ S = \frac{7\pi}{3} \ }\checkmark[/tex]