If has exactly 3 solutions in the interval then the roots of the equation belong to, JEE Main Previous Year Trigonometry Questions and Solutions - Solution of Triangles-2024

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Solution

Q.
If $2 \sin^{3} x + \sin 2 x \cos x + 4 \sin x - 4 = 0$ has exactly 3 solutions in the interval $[0, \frac{n π}{2}], n \in N,$ then the roots of the equation $x^{2} + n x + (n - 3) = 0$ belong to [2024]

1 $(0, \infty)$

2 $(- \infty, 0)$

3 $(- \frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2})$

4 $Z$

Ans.
(2)

$2 \sin^{3} x + \sin 2 x \cos x + 4 \sin x - 4 = 0$

$\Rightarrow 2 \sin^{3} x + 2 \sin x \cos^{2} x + 4 \sin x - 4 = 0$

$\Rightarrow 2 \sin x (\sin^{2} x + \cos^{2} x) + 4 \sin x = 4 \Rightarrow \sin x = \frac{4}{6} = \frac{2}{3}$

Clearly, from the graph, the given equation has exactly 3 solutions in $[0, \frac{5 π}{2}] \Rightarrow x = 5$

So, we have equation, $x^{2} + 5 x + 2 = 0$

$\Rightarrow x = \frac{- 5 + \sqrt{25 - 8}}{2} \Rightarrow x = \frac{- 5 + \sqrt{17}}{2} \in (- \infty, 0)$

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