Question 445384
For convenience substitute {{{theta=x}}} and modify the trigonometric equation:

{{{cos2x=cosx}}}, since {{{cos2x=(cosx)^2-(sinx)^2}}} and 

{{{(sinx)^2=1-(cosx)^2}}}, substituting we get:{{{(cosx)^2-(sinx)^2=cosx}}} =>

{{{(cosx)^2-1+(cosx)^2=cosx}}} => {{{2(cosx)^2-cosx-1=0}}}. In the last equation

substitute {{{cosx=y}}} and get the quadratic equation: {{{2y^2-y-1=0}}},

Solving this equation {{{2y^2-y-1=(2y+1)(y-1)=0}}} we find y=1 and y=-1/2.

Trigonometric equation {{{cos2x=cosx}}} is equivalent with two new equations:

{{{cosx=-1}}} and {{{cosx=-1/2}}}. Solving these equations we have;

{{{cosx=-1}}} => {{{x=180}}}degree and {{{cosx=-1/2}}} => {{{x=120}}}, and 

{{{x=240}}}degree.