How do you find the point of intersection of cosx=cos2x?

cosx = 2cos^2 x - 1

2cos^2 x - cosx - 1 = 0
(2cosx + 1)(cosx - 1) = 0
cosx = -1/2 or cosx = 1
x = 120° or 240° or x = 0 or x = 360°

Those are not "points of intersection" , they are solutions to your equation.
You can't find points of intersection, since you don't have y variable in your equation.

To find the point of intersection of the two equations cos(x) = cos(2x), we need to solve for x.

First, let's simplify the equation cos(2x). By applying the double-angle identity for cosine, we can rewrite it as cos(2x) = 2cos^2(x) - 1.

Now, we have the equation cos(x) = 2cos^2(x) - 1.

Rearranging the terms, we get 2cos^2(x) - cos(x) - 1 = 0.

This is now a quadratic equation in terms of cos(x). Let's substitute cos(x) with a variable, let's say 't'. Then the equation becomes 2t^2 - t - 1 = 0.

To solve this quadratic equation, you can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a.

For our equation 2t^2 - t - 1 = 0, where a = 2, b = -1, and c = -1, the quadratic formula becomes t = (-(-1) ± √((-1)^2 - 4(2)(-1))) / (2(2)).

Simplifying, we have t = (1 ± √(1 + 8))/ 4.

Now, we have two possible values for t. Plugging these values into cos(x) = t will give us the corresponding values for x.

Let's solve for each possible value:

When t = (1 + √(1 + 8))/ 4:
cos(x) = (1 + √9)/4
cos(x) = 1/2

Taking the inverse cosine (arccos) of both sides:
x = arccos(1/2)
x = π/3 or x = 2π/3

When t = (1 - √(1 + 8))/ 4:
cos(x) = (1 - √9)/4
cos(x) = -1

Taking the inverse cosine (arccos) of both sides:
x = arccos(-1)
x = π

Therefore, the points of intersection of cos(x) = cos(2x) are x = π/3, x = 2π/3, and x = π.

So, the point of intersection is (π/3, cos(π/3)), (2π/3, cos(2π/3)), and (π, cos(π)).