$\sin 2x = 1/2, 2x = 150, 390$. But why not 30? $0<x<360$
So $\sin(2x) = \sin (30 + 360)$ or $\sin(2x) = \sin (180-30)$.
It derives from 
\begin{align*}
\sin x - \cos x & =\frac{1}{\sqrt2}\\
\sin^2 x + \cos^2 x -2\sin x \cos x & = \frac{1}{2}\\
2\sin x \cos x & = \frac{1}{2}
\end{align*}
But why not $\sin (2x) = \sin (30)$ ? So $x = 15$.
The answer only $x = 75$ and $195$.

We have that
$$\sin 2x = \frac12=\sin 30° \iff 2x=30°+k360° \quad \lor \quad2x=180°-30°+k360°$$
therefore all the solutions are in the form


*

*$x=15°+k180°=15°,15°\pm 180°,15°\pm360°,\ldots$ 

*$x=75°+k180°=75°,75°\pm 180°,75°\pm360°,\ldots$
and for the range $0°\le x <360°$ the solutions are $15°$, $75°$, $195°$, $255°$.
