If 2 ^ 3 x + 2xcosx + 4 x - 4 = 0 has exactly 3 solutions in the...

Given: $2\sin^{3}x + \sin 2xcosx + 4\sin x - 4 = 0$

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

$⇒ 2\sin^{3}x + 2\sin x \cdot \left(1 - \sin^{2}x\right) + 4\sin x - 4 = 0$

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

$⇒ 6\sin x - 4 = 0$

$⇒ \sin x = \dfrac{2}{3}$

Now, for exactly three solution we get,

$⇒ n = 5$ (in the given interval)

So, ${x}^{2} + nx + \left(n - 3\right) = 0$

$⇒ {x}^{2} + 5x + 2 = 0$

$⇒ x = \dfrac{- 5 \pm \sqrt{17}}{2}$

So, the required interval is $( - \infty , 0 )$.