Question 729404
In the first quadrant, we know that {{{cos(60^o)=1/2}}} (using degrees), or {{{cos(pi/3)=1/2}}} if we prefer radians.
In the fourth quadrant, we also know that {{{cos(-60^o)=1/2}}} (using degrees), or {{{cos(-pi/3)=1/2}}} if we prefer radians.
Angles in the second and third quadrants have negative cosines, so no solutions there.
For angles co-terminal with those, we can add multiples of {{{360^o}}} , or {{{2pi}}} if we prefer radians to degrees.
All of those solutions can be written in one formula as
{{{2k*pi +- pi/3}}} if we prefer radians, or {{{k*360^o +- 60^o}}} if we prefer degrees.