Let S = \ x ∈ [-π, π] : x ( x + x) = a, a ∈ ℤ\ . Then n(S) is equal...

Given the equation: $\sin x (\sin x + \cos x) = a$

$\Rightarrow \sin^2 x + \sin x \cos x = a$

Using the double angle formulas $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ and $\sin x \cos x = \dfrac{\sin 2x}{2}$, we get:

$\dfrac{1 - \cos 2x}{2} + \dfrac{\sin 2x}{2} = a$

$\Rightarrow \sin 2x - \cos 2x = 2a - 1$

We know that the range of $A\sin \theta + B\cos \theta$ is $[-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}]$. Thus, the range of $\sin 2x - \cos 2x$ is $[-\sqrt{2}, \sqrt{2}]$.

$\Rightarrow -\sqrt{2} \le 2a - 1 \le \sqrt{2}$

$\Rightarrow 1 - 1.414 \le 2a \le 1 + 1.414$

$\Rightarrow -0.414 \le 2a \le 2.414$

$\Rightarrow -0.207 \le a \le 1.207$

Since $a \in \mathbb{Z}$, the possible values for $a$ are $0$ and $1$.

Case 1: $a = 0$
$\sin x (\sin x + \cos x) = 0$
$\Rightarrow \sin x = 0$ or $\sin x + \cos x = 0$
For $x \in [-\pi, \pi]$, $\sin x = 0 \Rightarrow x \in \{-\pi, 0, \pi\}$ (3 solutions)
$\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x \in \{-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\}$ (2 solutions)
Total solutions for $a = 0$ is $5$.

Case 2: $a = 1$
$\sin^2 x + \sin x \cos x = 1$
$\Rightarrow \sin x \cos x = 1 - \sin^2 x$
$\Rightarrow \sin x \cos x = \cos^2 x$
$\Rightarrow \cos x (\sin x - \cos x) = 0$
$\Rightarrow \cos x = 0$ or $\sin x - \cos x = 0$
For $x \in [-\pi, \pi]$, $\cos x = 0 \Rightarrow x \in \{-\dfrac{\pi}{2}, \dfrac{\pi}{2}\}$ (2 solutions)
$\sin x - \cos x = 0 \Rightarrow \tan x = 1 \Rightarrow x \in \{-\dfrac{3\pi}{4}, \dfrac{\pi}{4}\}$ (2 solutions)
Total solutions for $a = 1$ is $4$.

The total number of elements in the set $S$ is $n(S) = 5 + 4 = 9$.

Answer: $9$