Finding point(s) of intersection between $f(x)= \cos (2x)$ and $g(x)=- \cos x$

Question

Find the point(s) of intersection between $f(x)= \cos (2x)$ and $g(x)=- \cos x$, over interval $(0,2 \pi)$. Verify by graphing.

Answer 1

Here is a start of a solution... the graph:

Answer 2

All we need to do is set them equal to each other! So $\cos(2x)=-\cos(x)$. Then rearrange to $\cos(2x)+\cos(x)=0$ then change that to $2\cos^2(x)+\cos(x)-1=0$. Which is a quadratic equation which we get $\cos(x)= -1$ and $\cos(x)= \frac{1}{2}$.

So we get the general solutions $x=2n\pi+ \pi$, $x=\frac{5\pi}{3}+2n\pi$ and $x=\frac{\pi}{3}+2n\pi$

As for graphing you can use Desmos, or try and plot these graphs yourself and see where they intersect!

Edit: Graph for interest which backs up our claim!

Answer 3

Using the identity $\cos p+\cos q=2\cos(\frac{p+q}2)\cos(\frac{p-q}2)$,

$$\cos 2x+\cos x=2\cos(\frac32x)\cos(\frac12x)$$

This is zero iff $\frac32x=\frac\pi2+k\pi$ or $\frac12x=\frac\pi2+k\pi$, that is if $x=\frac\pi3+\frac{2k\pi}3$ or $x=\pi+2k\pi$, for $k\in\Bbb Z$.

You want the values in $(0,2\pi)$, so

$$x\in\{\frac\pi3,\pi,\frac{5\pi}{3}\}$$