How to calculate the length of the negative parts of $ \cos \alpha x-\cos 2 \alpha x $?

Question

I have this periodic function $$ f(x) = \cos (\alpha x)-\cos (2 \alpha x) $$ defined for $x>0$ and for $\alpha>0$.

How can I find the length of the graph of $f$ along one of its period when $f(x) < 0$?

Here I have plotted the function:

$\hspace{3.5cm}$

Answer 1

Too long for a comment:

The function is periodic in $x$ with period $T = -\frac{2\pi}{a}$.

Find the $x$ for which $f(x) < 0$ that are for example $x_1:=\frac{2\pi}{3a} \le x \le \frac{4\pi}{3a}=:x_2$. Then using the usual formula for evaluating the lenght of a curve you get:

$$L = \int_{x_1}^{x_2} \sqrt{1+f'(x) ^ 2}dx = \dots$$

[I used Wolfram for the calculations in an hurry]