Aus Online Mathematik Brückenkurs 1

First, we observe from the unit circle that the equation has two solutions for \displaystyle 0^{\circ}\le 3x\le 360^{\circ}\,,

\displaystyle 3x = 15^{\circ}\qquad\text{and}\qquad 3x = 180^{\circ} - 15^{\circ} = 165^{\circ}\,\textrm{.}

This means that all of the equation's solutions are

\displaystyle 3x = 15^{\circ} + n\cdot 360^{\circ}\qquad\text{and}\qquad 3x = 165^{\circ} + n\cdot 360^{\circ}\,,

for all integers n, i.e.

\displaystyle x = 5^{\circ} + n\cdot 120^{\circ}\qquad\text{and}\qquad x = 55^{\circ} + n\cdot 120^{\circ}\,\textrm{.}