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(D) Given the equation $x^2 + 6x + 12 + 3 \sin(a + b\alpha) = 0$. We can rewrite this as $x^2 + 6x + 9 + 3 + 3 \sin(a + b\alpha) = 0$. $(x + 3)^2 + 3(

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(D) Given the equation $x^2 + 6x + 12 + 3 \sin(a + b\alpha) = 0$. We can rewrite this as $x^2 + 6x + 9 + 3 + 3 \sin(a + b\alpha) = 0$. $(x + 3)^2 + 3(1 + \sin(a + b\alpha)) = 0$. Since $(x + 3)^2 \ge 0$ and $1 + \sin(a + b\alpha) \ge 0$ (because $-1 \le \sin 	heta \le 1$),the sum of two non-negative terms can be zero only if each term is zero. Therefore,$(x + 3)^2 = 0 \implies x = -3$ and $1 + \sin(a + b\alpha) = 0$. This implies $\sin(a + b\alpha) = -1$. For the least positive value of $	heta = a + b\alpha$,we have $\sin 	heta = -1$,which gives $	heta = \frac{3\pi}{2}$. We need to find $\cos(a + b\alpha) = \cos(\frac{3\pi}{2}) = 0$.

If $a, b$ are real numbers and $\alpha$ is a real root of $x^2 + 6x + 12 + 3 \sin(a + b\alpha) = 0$,then the value of $\cos(a + b\alpha)$ for the least positive value of $a + b\alpha$ is

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