If 2 sin^3 x + sin 2x cos x + 4 sin x - 4 = 0 | vedclass

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(B) Given equation: $2 \sin^3 x + (2 \sin x \cos x) \cos x + 4 \sin x - 4 = 0$$2 \sin^3 x + 2 \sin x \cos^2 x + 4 \sin x - 4 = 0$Substitute $\cos^2 x

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$(-\infty, 0)$

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(B) Given equation: $2 \sin^3 x + (2 \sin x \cos x) \cos x + 4 \sin x - 4 = 0$$2 \sin^3 x + 2 \sin x \cos^2 x + 4 \sin x - 4 = 0$Substitute $\cos^2 x = 1 - \sin^2 x$:$2 \sin^3 x + 2 \sin x (1 - \sin^2 x) + 4 \sin x - 4 = 0$$2 \sin^3 x + 2 \sin x - 2 \sin^3 x + 4 \sin x - 4 = 0$$6 \sin x = 4 \implies \sin x = \frac{2}{3}$Since $\sin x = \frac{2}{3} \approx 0.66$,in the interval $[0, \frac{n \pi}{2}]$,we need exactly $3$ solutions.For $n=1$,interval $[0, \pi/2]$,$\sin x = 2/3$ has $1$ solution.For $n=2$,interval $[0, \pi]$,$\sin x = 2/3$ has $2$ solutions.For $n=3$,interval $[0, 3\pi/2]$,$\sin x = 2/3$ has $2$ solutions.For $n=4$,interval $[0, 2\pi]$,$\sin x = 2/3$ has $2$ solutions.For $n=5$,interval $[0, 5\pi/2]$,$\sin x = 2/3$ has $3$ solutions.Thus,$n = 5$.The quadratic equation is $x^2 + 5x + 2 = 0$.Roots are $x = \frac{-5 \pm \sqrt{25 - 8}}{2} = \frac{-5 \pm \sqrt{17}}{2}$.Both roots are negative,so they belong to $(-\infty, 0)$.

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

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