If 2sin3x + sin 2x cos x + 4sinx - 4 = 0 has exactly 3 solutions in the interval \(\left[0, \frac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}\), then the roots of the equation x2 + nx + (n - 3) = 0 belong to:

1
(0, ∞)
2
(-∞, 0)
3
\(\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)\)
4
Z

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