Решите,пжл : sin(φ-π/3)=1/2

$Sin( \alpha - \frac{ \pi }{3})= \frac{1}{2} \\\\ \alpha - \frac{ \pi }{3}=arcSin \frac{1}{2} +2 \pi n\\\\ \alpha - \frac{ \pi }{3}= \frac{ \pi }{6} +2 \pi n\\\\ \alpha _{1}= \frac{ \pi }{3} + \frac{ \pi }{6}+ 2\pi n\\\\ \alpha _{1} = \frac{ \pi }{2}+2 \pi n \\\\\\ \alpha - \frac{ \pi }{3}= \pi -arcSin \frac{1}{2}+2 \pi n\\\\ \alpha - \frac{ \pi }{3}= \pi - \frac{ \pi }{6} +2 \pi n \\\\ \alpha - \frac{ \pi }{3}= \frac{5 \pi }{6} +2 \pi n\\\\ \alpha _{2}= \frac{ \pi }{3}+ \frac{5 \pi }{6}+ 2 \pi n\\\\ \alpha _{2}= \frac{7 \pi }{6} +2 \pi n$ n∈z