Losing solutions in $\cos{x}+\cos{2x}=0$

Question

I'm trying to solve $\cos x=-\cos2x$, $-\pi<x\leq\pi$, WITHOUT USING the double angle formulae, here's my current working: $$\begin{align*}\cos x&=-\cos2x\\ \cos (x\pm2k\pi)&=\cos (2x\pm(2k+1)\pi)\\ \implies x\pm2k\pi&=2x\pm(2k+1)\pi\\ x&=2x\pm\pi\\ x&=\pm\pi\\ \therefore x&=\pi\text{ , in the given range of $x$}\end{align*}$$

Graphically it's obvious I've missed two other solutions, $x=\pm\frac{\pi}{3}$, however I do not know why because I have not, to my knowledge, carried out any illegal operations such as dividing by zero.

Could someone point out my mistake please and give an edited solution? Thanks!

Answer 1

Use that

$$\cos x=-\cos2x \iff \cos x=\cos(\pi-2x)$$

and that

$$\cos \alpha = \cos \theta \iff \alpha = \theta +2k\pi \, \lor \, \alpha = -\theta +2k\pi \quad k\in \mathbb{Z}$$

Refer also to the related

• Solving $\cos(3x) = \cos(2x)$

Answer 2

$$\cos(x+\pi)=\cos(2x)\iff x+\pi=2k\pi\pm2x.$$

Hence

$$x=(2k+1)\pi\lor x=(2k+1)\frac\pi3.$$