Aus Online Mathematik Brückenkurs 1

First, we observe from the unit circle that the equation has two solutions for \displaystyle 0^{\circ }\le \text{3}x\le \text{36}0^{\circ },

\displaystyle 3x=15^{\circ } and \displaystyle 3x=180^{\circ }-15^{\circ }=165^{\circ }

This means that all of the equation's solutions are

\displaystyle 3x=15^{\circ }+n\centerdot 360^{\circ } and \displaystyle 3x=165^{\circ }+n\centerdot 360^{\circ }

for all integers \displaystyle n, i.e.

\displaystyle x=5^{\circ }+n\centerdot 120^{\circ } and \displaystyle x=55^{\circ }+n\centerdot 120^{\circ }