If S = \ θ ∈ [-π, π] : θ (5θ)/(2) = 7θ (7θ)/(2)\ , then n(S) is...

Given equation: $\cos\theta \cos\dfrac{5\theta}{2} = \cos 7\theta \cos\dfrac{7\theta}{2}$

Multiplying both sides by $2$:

$2\cos\theta \cos\dfrac{5\theta}{2} = 2\cos 7\theta \cos\dfrac{7\theta}{2}$

Using the identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:

$\cos\left(\theta + \dfrac{5\theta}{2}\right) + \cos\left(\theta - \dfrac{5\theta}{2}\right) = \cos\left(7\theta + \dfrac{7\theta}{2}\right) + \cos\left(7\theta - \dfrac{7\theta}{2}\right)$

$\cos\left(\dfrac{7\theta}{2}\right) + \cos\left(\dfrac{-3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right) + \cos\left(\dfrac{7\theta}{2}\right)$

Since $\cos(-x) = \cos x$, we get:

$\cos\left(\dfrac{7\theta}{2}\right) + \cos\left(\dfrac{3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right) + \cos\left(\dfrac{7\theta}{2}\right)$

$\cos\left(\dfrac{3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right)$

$\cos\left(\dfrac{21\theta}{2}\right) - \cos\left(\dfrac{3\theta}{2}\right) = 0$

Using the identity $\cos C - \cos D = -2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right)$:

$-2\sin\left(\dfrac{\dfrac{21\theta}{2} + \dfrac{3\theta}{2}}{2}\right)\sin\left(\dfrac{\dfrac{21\theta}{2} - \dfrac{3\theta}{2}}{2}\right) = 0$

$-2\sin(6\theta)\sin\left(\dfrac{9\theta}{2}\right) = 0$

This gives $\sin(6\theta) = 0$ or $\sin\left(\dfrac{9\theta}{2}\right) = 0$.

Case 1: $\sin(6\theta) = 0$

$6\theta = n\pi \Rightarrow \theta = \dfrac{n\pi}{6}$ for $n \in \mathbb{Z}$.

Since $\theta \in [-\pi, \pi]$, we have $-\pi \le \dfrac{n\pi}{6} \le \pi \Rightarrow -6 \le n \le 6$.

The possible values for $n$ are $\{-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$.

This gives $13$ solutions.

Case 2: $\sin\left(\dfrac{9\theta}{2}\right) = 0$

$\dfrac{9\theta}{2} = m\pi \Rightarrow \theta = \dfrac{2m\pi}{9}$ for $m \in \mathbb{Z}$.

Since $\theta \in [-\pi, \pi]$, we have $-\pi \le \dfrac{2m\pi}{9} \le \pi \Rightarrow -4.5 \le m \le 4.5$.

The possible values for $m$ are $\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$.

This gives $9$ solutions.

To find common solutions, we equate the two sets of solutions:

$\dfrac{n\pi}{6} = \dfrac{2m\pi}{9} \Rightarrow 3n = 4m$

For $m \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$, $m$ must be a multiple of $3$. The valid values for $m$ are $-3, 0, 3$, which correspond to $n = -4, 0, 4$.

Thus, there are $3$ common solutions.

Total number of unique solutions $n(S) = 13 + 9 - 3 = 19$.

Answer: $19$