Question

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

• A. \( (0, \infty) \)
• B. \( (-\infty, 0) \)
• C. \( \left( -\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2} \right) \)
• D. \( \mathbb{Z} \)

Answer 1

The Correct Option is B

To solve the problem, we need to determine the interval of the roots of the quadratic equation \( x^2 + nx + (n - 3) = 0 \). The equation \( 2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0 \) gives exactly 3 solutions in the interval \( \left[ 0, \frac{n \pi}{2} \right] \). Let's follow the step-by-step solution:

1. First, identify the nature of the trigonometric equation \( 2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0 \).
2. To solve for \( n \), we observe that within the interval \( \left[ 0, \frac{n \pi}{2} \right] \), there are 3 solutions, which suggests that one complete period of the sine function \(\sin x\) contributes to the solutions. Notably, the sine function completes a full period at \(\pi\).
3. Thus, for 3 solutions, \( n \) must equal 1.5, positioning \(\frac{3\pi}{2}\) as the end of the interval.
4. With \( n = 3 \), substitute into the given quadratic equation:
5. Factor out the common term:
6. The roots are \( x = 0 \) and \( x = -3 \).
7. Examine the intervals provided in the options. With the roots \( x = 0 \) and \( x = -3 \), both roots being less than or equal to zero means the roots belong to the interval \((-∞, 0]\).
8. Therefore, the correct answer is the interval option:
   • \((-\infty, 0)\)

In conclusion, through identifying functional transferences and factorizations, the answer to the problem is that the roots of the given quadratic equation belong to the interval \((-∞, 0)\).