Number of Solutions for Trigonometric Equation with Integer Constant

To find the number of elements in the set $S = \{x \in [-\pi, \pi] : \sin x (\sin x + \cos x) = a, a \in \mathbb{Z}\}$, we first simplify the given trigonometric equation.

Let $f(x) = \sin x (\sin x + \cos x)$. Expanding this, we get: $f(x) = \sin^2 x + \sin x \cos x$

Using the trigonometric identities $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\sin x \cos x = \frac{\sin 2x}{2}$, we can rewrite $f(x)$ as: $f(x) = \frac{1 - \cos 2x}{2} + \frac{\sin 2x}{2} = \frac{1}{2} + \frac{1}{2}(\sin 2x - \cos 2x)$

Using the linear combination identity $\sin \theta - \cos \theta = \sqrt{2} \sin\left(\theta - \frac{\pi}{4}\right)$, we obtain: $f(x) = \frac{1}{2} + \frac{1}{\sqrt{2}} \sin\left(2x - \frac{\pi}{4}\right)$

Step 1: Range of $f(x)$ and possible values of integer $a$

Since the sine function takes values in $[-1, 1]$, the range of $f(x)$ is given by: $f(x) \in \left[\frac{1}{2} - \frac{1}{\sqrt{2}}, \frac{1}{2} + \frac{1}{\sqrt{2}}\right]$

Calculating the numerical bounds: $f(x) \in \left[\frac{1 - \sqrt{2}}{2}, \frac{1 + \sqrt{2}}{2}\right] \approx [-0.207, 1.207]$

Since $a \in \mathbb{Z}$, the only integers lying in this range are: $a = 0 \quad \text{and} \quad a = 1$

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Step 2: Finding solutions for $a = 0$

Setting $f(x) = 0$: $\frac{1}{2} + \frac{1}{\sqrt{2}} \sin\left(2x - \frac{\pi}{4}\right) = 0 \implies \sin\left(2x - \frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$

Let $y = 2x - \frac{\pi}{4}$. Since $x \in [-\pi, \pi]$, we have: $2x \in [-2\pi, 2\pi] \implies y \in \left[-2\pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4}\right] = \left[-\frac{9\pi}{4}, \frac{7\pi}{4}\right]$

Now we look for values of $y \in \left[-\frac{9\pi}{4}, \frac{7\pi}{4}\right]$ such that $\sin y = -\frac{1}{\sqrt{2}}$: $y \in \left\{-\frac{9\pi}{4}, -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\right\}$

Converting back to $x = \frac{1}{2}\left(y + \frac{\pi}{4}\right)$:

• $y = -\frac{9\pi}{4} \implies x = -\pi$
• $y = -\frac{3\pi}{4} \implies x = -\frac{\pi}{4}$
• $y = -\frac{\pi}{4} \implies x = 0$
• $y = \frac{5\pi}{4} \implies x = \frac{3\pi}{4}$
• $y = \frac{7\pi}{4} \implies x = \pi$

Thus, there are $5$ solutions for $a = 0$.

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Step 3: Finding solutions for $a = 1$

Setting $f(x) = 1$: $\frac{1}{2} + \frac{1}{\sqrt{2}} \sin\left(2x - \frac{\pi}{4}\right) = 1 \implies \sin\left(2x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

For $y \in \left[-\frac{9\pi}{4}, \frac{7\pi}{4}\right]$, the solutions to $\sin y = \frac{1}{\sqrt{2}}$ are: $y \in \left\{-\frac{7\pi}{4}, -\frac{5\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}\right\}$

Converting back to $x = \frac{1}{2}\left(y + \frac{\pi}{4}\right)$:

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

Thus, there are $4$ solutions for $a = 1$.

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Step 4: Total Number of Solutions

Combining all solutions: $S = \left\{-\pi, -\frac{3\pi}{4}, -\frac{\pi}{2}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\right\}$

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

Hence, the correct option is D.