JEE Main 202430 Jan, Morning shiftMathematicsEasyTrigonometry118× since 2002If 2sin3x+sin2xcosx+4sinx−4=02 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=02sin3x+sin2xcosx+4sinx−4=0 has exactly 3 solutions in the interval [0,nπ2],n∈N\left[0, \frac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}[0,2nπ],n∈N, then the roots of the equation x2+nx+(n−3)=0x^2+\mathrm{n} x+(\mathrm{n}-3)=0x2+nx+(n−3)=0 belong to :AOption A. (0,∞)(0, \infty)(0,∞)BOption B. ZCOption C. (−172,172)\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)(−217,217)DOption D. (−∞,0)(-\infty, 0)(−∞,0)CheckSkip · show solution
