**cos x - cos 2x = 0**

So far, I've subbed in 2cos\({^2}\)(x) - 1; **cos x - 2cos\({^2}\)x +1 = 0**

I'm not sure what I can do next?

Thank you! :)

Guest Feb 22, 2019

#1**+1 **

\(\cos(x) - \cos(2x) = 0\\ \cos(x) = \cos(2x)\\ \cos(x) = 2\cos^2(x) -1\\ 2\cos^2(x) - \cos(x) - 1 = 0\)

\(u = \cos(x)\\ 2u^2 - u - 1 = 0\\ (2u+1 )(u -1) = 0\\ u = 1,~-\dfrac 1 2\)

\(\cos(x) = 1 \Rightarrow x = 2k\pi, ~k \in \mathbb{Z}\\ \cos(x) = -\dfrac 1 2 \Rightarrow x = \dfrac{2\pi}{3}+2k\pi,~\dfrac{4\pi}{3}+2k\pi,~k\in \mathbb{Z}\)

\(x = \left \{0,~\dfrac{2\pi}{3},~\dfrac{4\pi}{3}\right\} + 2\pi k,~k \in \mathbb{Z}\)

.Rom Feb 22, 2019