Question:

Step 1: Analyze the constraint on the sum of sines. We are given that $\sin\alpha + \sin\beta + \sin\gamma = -3$ where $\alpha, \beta, \gamma \in (0, 2\pi)$. Since the sine function is bounded, we know that $-1 \leq \sin\theta \leq 1$ for all $\theta$. This means each individual sine value is at most $1$ and at least $-1$. Step 2: Determine the individual sine values. For the sum of three terms, each bounded by $-1$, to equal exactly $-3$, we must have: $$\sin\alpha = \sin\beta = \sin\gamma = -1$$ This is the only way three numbers, each between $-1$ and $1$, can sum to $-3$. Step 3: Find the values of $\alpha$, $\beta$, and $\gamma$. We need to find which values in $(0, 2\pi)$ satisfy $\sin\theta = -1$. The sine function equals $-1$ at: $$\theta = \frac{3\pi}{2}$$ Therefore: $$\alpha = \beta = \gamma = \frac{3\pi}{2}$$ Step 4: Calculate $\cos 2\alpha$. $$\cos 2\alpha = \cos\left(2 \cdot \frac{3\pi}{2}\right) = \cos(3\pi)$$ Since $\cos(3\pi) = \cos(\pi) = -1$: $$\cos 2\alpha = -1$$ Step 5: Calculate $\cos 4\beta$. $$\cos 4\beta = \cos\left(4 \cdot \frac{3\pi}{2}\right) = \cos(6\pi)$$ Since $\cos(6\pi) = \cos(0) = 1$: $$\cos 4\beta = 1$$ Step 6: Calculate $\cos 6\gamma$. $$\cos 6\gamma = \cos\left(6 \cdot \frac{3\pi}{2}\right) = \cos(9\pi)$$ Since $\cos(9\pi) = \cos(\pi) = -1$: $$\cos 6\gamma = -1$$ Step 7: Find the sum. $$\cos 2\alpha + \cos 4\beta + \cos 6\gamma = (-1) + 1 + (-1) = -1$$ The final answer is $\boxed{-1}$, which corresponds to **Option 1**.