Question 732671
We know that {{{cos(60^o)=1/2}}} (or {{{cos(pi/3)=1/2}}} if measuring angles in radians).
Negative cosines happen only in the second and third quadrants.
{{{cos(180^o-60^o)=cos(120^o) =-cos(60^o)=-1/2}}}
(or {{{cos(pi-pi/3)=cos(2pi/3)=-cos(pi/3)=-1/2}}} in radians) shows us one of the solutions.
An angle with the same cosine in the third quadrant is {{{-120^o}}} (or {{{-2pi/3}}} in radians and that is another solution.
The other solutions are all angles coterminal with those two angles.
They can all be expressed as
{{{360^o +- k*120^o}}} or {{{2pi +- 2k*pi/3}}} for any integer {{{k}}}.