We are asked to find the number of solutions, $n(S)$, for the equation $\sin x (\sin x + \cos x) = a$, where $x \in [-\pi, \pi]$ and $a \in \mathbf{Z}$.

Solving the Trigonometric Equation

1. Rewrite the equation:

   Expand the given equation:

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

   Use trigonometric identities $\sin^2 x = \frac{1 - \cos(2x)}{2}$ and $\sin x \cos x = \frac{\sin(2x)}{2}$:

   $ \frac{1 - \cos(2x)}{2} + \frac{\sin(2x)}{2} = a $

   Multiply by 2:

   $ 1 - \cos(2x) + \sin(2x) = 2a $

   Rearrange the terms:

   $ \sin(2x) - \cos(2x) = 2a - 1 $

2. Convert to $R\sin(\theta - \alpha)$ form:

   The expression $\sin(2x) - \cos(2x)$ can be written as $R\sin(2x - \alpha)$, where $R = \sqrt{1^2 + (-1)^2} = \sqrt{2}$ and $\alpha$ is such that $\cos \alpha = \frac{1}{\sqrt{2}}$ and $\sin \alpha = \frac{1}{\sqrt{2}}$. Thus, $\alpha = \frac{\pi}{4}$.

   So the equation becomes:

   $ \sqrt{2} \sin\left(2x - \frac{\pi}{4}\right) = 2a - 1 $

   Isolate the sine term:

   $ \sin\left(2x - \frac{\pi}{4}\right) = \frac{2a - 1}{\sqrt{2}} $

3. Determine possible integer values for $a$:

   The range of the sine function is $[-1, 1]$. Therefore:

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

   Multiply by $\sqrt{2}$:

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

   Add 1:

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

   Divide by 2:

   $ \frac{1 - \sqrt{2}}{2} \le a \le \frac{1 + \sqrt{2}}{2} $

   Using $\sqrt{2} \approx 1.414$:

   $ \frac{1 - 1.414}{2} \le a \le \frac{1 + 1.414}{2} $

   $ \frac{-0.414}{2} \le a \le \frac{2.414}{2} $

   $ -0.207 \le a \le 1.207 $

   Since $a$ must be an integer ($a \in \mathbf{Z}$), the possible values for $a$ are $0$ and $1$.

4. Find solutions for $a=0$:

   Substitute $a=0$ into the equation:

   $ \sin\left(2x - \frac{\pi}{4}\right) = \frac{2(0) - 1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} $

   Let $y = 2x - \frac{\pi}{4}$. Since $x \in [-\pi, \pi]$, the range for $2x$ is $[-2\pi, 2\pi]$, and the range for $y$ is $[-2\pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4}] = [-\frac{9\pi}{4}, \frac{7\pi}{4}]$.

   The values of $y$ in the range $[-\frac{9\pi}{4}, \frac{7\pi}{4}]$ for which $\sin y = -\frac{1}{\sqrt{2}}$ are:

   $y = -\frac{9\pi}{4}, -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

   Now, find $x$ using $x = \frac{y + \pi/4}{2}$:

   • $y = -\frac{9\pi}{4} \implies x = \frac{-9\pi/4 + \pi/4}{2} = \frac{-8\pi/4}{2} = -\pi$
   • $y = -\frac{3\pi}{4} \implies x = \frac{-3\pi/4 + \pi/4}{2} = \frac{-2\pi/4}{2} = -\frac{\pi}{4}$
   • $y = -\frac{\pi}{4} \implies x = \frac{-\pi/4 + \pi/4}{2} = 0$
   • $y = \frac{5\pi}{4} \implies x = \frac{5\pi/4 + \pi/4}{2} = \frac{6\pi/4}{2} = \frac{3\pi}{4}$
   • $y = \frac{7\pi}{4} \implies x = \frac{7\pi/4 + \pi/4}{2} = \frac{8\pi/4}{2} = \pi$

   There are 5 solutions for $a=0$. These are $x = -\pi, -\frac{\pi}{4}, 0, \frac{3\pi}{4}, \pi$. All are within $[-\pi, \pi]$.

5. Find solutions for $a=1$:

   Substitute $a=1$ into the equation:

   $ \sin\left(2x - \frac{\pi}{4}\right) = \frac{2(1) - 1}{\sqrt{2}} = \frac{1}{\sqrt{2}} $

   Let $y = 2x - \frac{\pi}{4}$. The range for $y$ is $[-\frac{9\pi}{4}, \frac{7\pi}{4}]$.

   The values of $y$ in the range $[-\frac{9\pi}{4}, \frac{7\pi}{4}]$ for which $\sin y = \frac{1}{\sqrt{2}}$ are:

   $y = -\frac{7\pi}{4}, -\frac{5\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}$

   Now, find $x$ using $x = \frac{y + \pi/4}{2}$:

   • $y = -\frac{7\pi}{4} \implies x = \frac{-7\pi/4 + \pi/4}{2} = \frac{-6\pi/4}{2} = -\frac{3\pi}{4}$
   • $y = -\frac{5\pi}{4} \implies x = \frac{-5\pi/4 + \pi/4}{2} = \frac{-4\pi/4}{2} = -\frac{\pi}{2}$
   • $y = \frac{\pi}{4} \implies x = \frac{\pi/4 + \pi/4}{2} = \frac{2\pi/4}{2} = \frac{\pi}{4}$
   • $y = \frac{3\pi}{4} \implies x = \frac{3\pi/4 + \pi/4}{2} = \frac{4\pi/4}{2} = \frac{\pi}{2}$

   There are 4 solutions for $a=1$. These are $x = -\frac{3\pi}{4}, -\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{2}$. All are within $[-\pi, \pi]$.

6. Calculate the total number of solutions $n(S)$:

   Total solutions = (Solutions for $a=0$) + (Solutions for $a=1$)

   $ n(S) = 5 + 4 = 9 $

Therefore, the number of solutions $n(S)$ is 9.