3. Determine all positive values of the parameter $a$ for which the non-negative values of $x$ that satisfy the equation

$$
\cos ((8 a-3) x)=\cos ((14 a+5) x)
$$

form an increasing arithmetic progression.
Solution. The given equation will be written in the form

$$
\cos ((8 a-3) x)-\cos ((14 a+5) x)=0
$$

i.e.

$$
-2 \sin \frac{(8 a-3) x+(14 a+5) x}{2} \sin \frac{(8 a-3) x-(14 a+5) x}{2}=0
$$

Thus, the given equation is equivalent to the system of equations

$$
\left[\begin{array}{l}
(8 a-3) x+(14 a+5) x=2 k \pi \\
(8 a-3) x-(14 a+5) x=2 n \pi
\end{array}\right.
$$

or

$$
\left[\begin{array}{l}
(11 a+1) x=k \pi \\
(3 a+4) x=n \pi
\end{array}\right.
$$

Since $a>0$, it follows that $11 a+1>0$ and $3 a+4>0$. Thus $x=\frac{k \pi}{11 a+1}$ or $x=\frac{n \pi}{3 a+4}$. We denote $x_{k}=\frac{k \pi}{11 a+1}$ and $x_{n}=\frac{n \pi}{3 a+4}$. Since $x \geq 0$, we get that $n, k=0,1,2, \ldots$, so ( $x_{k}$ ) and ( $x_{n}$ ) are two increasing arithmetic progressions with first terms 0 and differences $d_{1}=\frac{\pi}{11 a+1}$ and $d_{2}=\frac{\pi}{3 a+4}$ respectively.

We will prove the following property: the numbers $x_{k}$ and $y_{k}$ form an arithmetic progression if and only if: for $d_{1} \leq d_{2}$ there exists $m \in \mathbb{N}$ such that $d_{2}=m d_{1}$, or for $d_{2} \leq d_{1}$ there exists $m \in \mathbb{N}$ such that $d_{1}=m d_{2}$.

Indeed, if $d_{1} \leq d_{2}$, then $d_{1}$ is the second term of the new arithmetic progression (the first is 0). Since $\frac{\pi}{11 a+1} \leq \frac{\pi}{3 a+4}$, it follows that the difference of the new arithmetic progression is $d_{1}$. But $d_{2}$ is also a term of this arithmetic progression, so there exists $m \in \mathbf{N}$ such that $d_{2}=m d_{1}$.

Conversely, let there exist $m \in \mathbb{N}$ such that $d_{2}=m d_{1}$. Then $x_{n}=d_{2} m(n-1)$. This means that every term of the progression ( $x_{n}$ ) is a term of the progression ( $x_{k}$ ), so the desired progression is $\left(x_{k}\right)$. Similarly, it is proved for $d_{2} \leq d_{1}$.

Furthermore, let $d_{1} \leq d_{2}$. Then there exists $m \in \mathbb{N}$ such that $d_{2}=m d_{1}$, i.e., $m \frac{\pi}{11 a+1}=\frac{\pi}{3 a+4}$, from which it follows that $a=\frac{4 m-1}{11-3 m}$. We have $m \in \mathbb{N}$, so for $4 m-1>0$, $a>0$, it follows that $11-3 m>0$, so $m \leq \frac{11}{3}$. Thus $m \in\{1,2,3\}$ and the corresponding values for $a$ are $\frac{3}{8}, \frac{7}{5}, \frac{11}{3}$.

Now let $d_{2} \leq d_{1}$. Then there exists $m \in \mathbb{N}$ such that $d_{1}=d_{2} m$, i.e., $\frac{\pi}{11 a+1}=m \frac{\pi}{3 a+4}$, from which $a=\frac{4-m}{11 m-3}$. For $m \in \mathbb{N}$ and $a>0$, we get, for similar reasons as before, $m \in\{1,2,3\}$. The corresponding values for $a$ are $\frac{3}{8}, \frac{2}{19}, \frac{1}{30}$.

Thus, $a \in\left\{\frac{3}{8}, \frac{2}{19}, \frac{1}{30}, \frac{7}{5}, \frac{11}{3}\right\}$.