JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 17)
If $$2 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=0$$ 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 $$x^2+\mathrm{n} x+(\mathrm{n}-3)=0$$ belong to :
$$(0, \infty)$$
Z
$$\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)$$
$$(-\infty, 0)$$
Explanation
$$\begin{aligned}
& 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0 \\
& 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \\
& 6 \sin x-4=0 \\
& \sin x=\frac{2}{3} \\
& \mathbf{n}=5 \text { (in the given interval) } \\
& x^2+5 x+2=0 \\
& x=\frac{-5 \pm \sqrt{17}}{2} \\
& \text { Required interval }(-\infty, 0)
\end{aligned}$$
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