$\sin \alpha =\frac{1}{2}$  

${\alpha }_1=\frac{\pi }{6}+2k\pi ={30}^o+k\cdot {360}^o\vee {\alpha }_2=\pi -\frac{\pi }{6}+2k\pi ={150}^o+k\cdot {360}^o$ ${\alpha }_1=\frac{\pi }{6}+2k\pi ={30}^o+k\cdot {360}^o\vee {\alpha }_2=\pi -\frac{\pi }{6}+2k\pi ={150}^o+k\cdot {360}^o$   

$k=5$  

${\alpha }_2={150}^o+{1800}^o={1950}^o$  

$\text{Odpowiedzˊ D.}$