Question:

Step 1: Substitute $t = \sin\alpha$ to simplify the expression. Since $\sin\alpha$ can take any value in $[-1, 1]$, let $t = \sin\alpha$ where $t \in [-1, 1]$ and $t \neq 1$ (to avoid division by zero). Also, note that $\cos^2\alpha = 1 - \sin^2\alpha = 1 - t^2$. Step 2: Simplify the numerator by substituting $t$ and $\cos^2\alpha$. The numerator is: $$\sin^3\alpha + 6\sin^2\alpha + \sin\alpha + 2\cos^2\alpha - 8$$ Substituting $\sin\alpha = t$ and $\cos^2\alpha = 1 - t^2$: $$t^3 + 6t^2 + t + 2(1-t^2) - 8$$ Expanding: $$t^3 + 6t^2 + t + 2 - 2t^2 - 8 = t^3 + 4t^2 + t - 6$$ Step 3: Factor the numerator polynomial. We need to factor $t^3 + 4t^2 + t - 6$. Testing $t = 1$: $$1 + 4 + 1 - 6 = 0$$ So $(t-1)$ is a factor. Using polynomial division: $$t^3 + 4t^2 + t - 6 = (t-1)(t^2 + 5t + 6)$$ Factoring the quadratic: $$t^2 + 5t + 6 = (t+2)(t+3)$$ Therefore: $$t^3 + 4t^2 + t - 6 = (t-1)(t+2)(t+3)$$ Step 4: Simplify the original expression by canceling the common factor. The expression becomes: $$\frac{(t-1)(t+2)(t+3)}{t-1} = (t+2)(t+3)$$ for $t \neq 1$. Expanding: $$(t+2)(t+3) = t^2 + 5t + 6$$ Step 5: Find the minimum of the quadratic function on the domain $[-1, 1)$. The quadratic $f(t) = t^2 + 5t + 6$ has its vertex at: $$t = -\frac{5}{2} = -2.5$$ Since the vertex is at $t = -2.5$, which lies outside the domain $[-1, 1)$, the function is monotonically increasing on $[-1, 1)$. Step 6: Evaluate the function at the left endpoint of the domain. The minimum value occurs at $t = -1$: $$f(-1) = (-1)^2 + 5(-1) + 6 = 1 - 5 + 6 = 2$$ Alternatively, using the factored form: $$f(-1) = (-1+2)(-1+3) = (1)(2) = 2$$ **The minimum value of the given expression is $\boxed{2}$, which corresponds to Option 2.**