решите уравнение:
а)3tg(x-π/3)+√3=0
б)sin⁡(x+π/3)=0

3tg(x - \dfrac{ \pi }{3}) + \sqrt{3} = 0 \\ \\ 3tg(x - \dfrac{ \pi }{3}) =- \sqrt{3} \\ \\ tg(x - \dfrac{ \pi }{3}) = - \dfrac{ \sqrt{3} }{3} \\ \\ x - \dfrac{ \pi }{3} = -\dfrac{ \pi}{6} + \pi n, \ n \in Z \\ \\ \boxed{x = \dfrac{ \pi }{6} + \pi n, \ n \in Z}sin(x + \dfrac{ \pi }{3}) = 0 \\ \\ x + \dfrac{\pi}{3} = \pi n, \ n \in Z \\ \\ \boxed{x = - \dfrac{ \pi }{3} + \pi n, \ n \in Z}