Trigonometric Equations Question 6
Question 6 - 2024 (30 Jan Shift 1)
If $2 \sin ^{3} x+\sin 2 x \cos x+4 \sin x-4=0$ has exactly 3 solutions in the interval $\left[0, \frac{n \pi}{2}\right], n \in N$, then the roots of the equation $x^{2}+n x+(n-3)=0$ belong to :
(1) $(0, \infty)$
(2) $(-\infty, 0)$
(3) $\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)$
(4) Z
Show Answer
Answer (2)
Solution
$2 \sin ^{3} x+2 \sin x \cdot \cos ^{2} x+4 \sin x-4=0$
$2 \sin ^{3} x+2 \sin x \cdot\left(1-\sin ^{2} x\right)+4 \sin x-4=0$
$6 \sin x-4=0$
$\sin x=\frac{2}{3}$
$\mathbf{n}=\mathbf{5}$ (in the given interval)
$x^{2}+5 x+2=0$
$x=\frac{-5 \pm \sqrt{17}}{2}$
Required interval $(-\infty, 0)$
