Question 445380
{{{cos(2THETA) = cos(THETA)}}}
Using the identity {{{cos(2THETA) = 2(cos(THETA))^2 - 1}}} we have
{{{2(cos(THETA))^2 - 1 = cos(THETA) -> 2(cos(THETA))^2 - cos(THETA) - 1 = 0}}}
This is a quadratic in {{{cos(THETA)}}}.  Let x = {{{cos(THETA)}}}.  Then we can write:
{{{2x^2 - x - 1 = 0}}}
Solve for x using the quadratic formula:
{{{x = (1 +- sqrt(1 + 8))/4}}}
This gives x = 1, -1/2
Since {{{x = cos(THETA)}}} we need to find values for {{{THETA}}} which give {{{cos(THETA) = 1}}}, {{{-1/2}}}
The inv. cos of 1 = 0 deg.
The inv. cos of -1/2 = 120 deg.
Since the {{{THETA}}} ranges from 0 to 360 deg. and since {{{2THETA}}} = 240 deg., we need to include 240 deg.
So the answers are {{{THETA = 0}}}deg, {{{THETA = 120}}}deg, {{{THETA = 240}}}deg