JEE Main 2024 — Mathematics Trigonometry

Given: $2{\mathrm{sin}}^{3}x+\mathrm{sin}2x\mathrm{cos}x+4\mathrm{sin}x-4=0$2sin3x+sin2xcosx+4sinx−4=0

$$\Rightarrow 2{\mathrm{sin}}^{3}x+2\mathrm{sin}x\cdot {\mathrm{cos}}^{2}x+4\mathrm{sin}x-4=0$$⇒2sin3x+2sinx⋅cos2x+4sinx−4=0

$$\Rightarrow 2{\mathrm{sin}}^{3}x+2\mathrm{sin}x\cdot (1-{\mathrm{sin}}^{2}x)+4\mathrm{sin}x-4=0$$⇒2sin3x+2sinx⋅(1−sin2x)+4sinx−4=0

$$\Rightarrow 2{\mathrm{sin}}^{3}x+2\mathrm{sin}x-2{\mathrm{sin}}^{3}x+4\mathrm{sin}x-4=0$$⇒2sin3x+2sinx−2sin3x+4sinx−4=0

$$\Rightarrow 6\mathrm{sin}x-4=0$$⇒6sinx−4=0

$$\Rightarrow \mathrm{sin}x=\frac{2}{3}$$⇒sinx=32​

Now, for exactly three solution we get,

$\Rightarrow n=5$⇒n=5 (in the given interval)

So, ${x}^{2}+nx+(n-3)=0$x2+nx+(n−3)=0

$$\Rightarrow {x}^{2}+5x+2=0$$⇒x2+5x+2=0

$$\Rightarrow x=\frac{-5\pm \sqrt{17}}{2}$$⇒x=2−5±17​​

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