cos(2x) = -cos(x)
2cos^(x) - 1= - cos(x)
2cos^2(x) + cos(x) - 1 = 0 factor
(2cos(x) - 1) (cos(x) + 1) = 0 and setting each factor to 0, we have
2cos(x) - 1 = 0
add 1 to both sides
2cos(x) = 1
divide both sides by 2
cos(x) = 1/2
And this happens at pi/3 + 2pi(n)
and at 5pi/3 + 2pi(n)
where n is an integer
cos(x) + 1 = 0
subtract 1 from each side
cos(x) = -1
and this happens at pi + 2pi(n)
where n is some integer
Here's a graph of the solutions........https://www.desmos.com/calculator/kietyhmst0