Question

How do you solve `cos2x=cos^2x` from `[-pi,pi]`?

Answer 1

Solve `cos 2x = cos^2 x`

Explanation:

Use trig identity: `cos 2x = 2cos^2 x - 1`

`2cos^2 x - 1 = cos^2 x --> cos ^2 x = 1 --> cos x = + 1 -> x = 0`

Between interval `(-pi, pi) -> x = 0`