The number of solutions of sin 2 x+ (2+2 x-x 2 ) sin x-3(x-1) 2 =0 , where -π leq x leq π , is

Mathematics · Functions

JEE Main 2024 — 5 April, Shift 2 — Question 23

The number of solutions of $\sin ^{2} x+\left(2+2 x-x^{2}\right) \sin x-3(x-1)^{2}=0$ , where $-\pi \leq \mathrm{x} \leq \pi$ , is

Numerical answer — enter this value.

$\quad \sin ^{2} \mathrm{x}-\left(\mathrm{x}^{2}-2 \mathrm{x}-2\right) \sin \mathrm{x}-3(\mathrm{x}-1)^{2}=0$

$\left.\sin ^{2} x-(x-1)^{2}\right) \sin x-3(x-1)^{2}=0$

roots : $\sin x=-3($ reject $)$ or $(x-1)^{2}$

$\Rightarrow \sin x=(x-1)^{2}$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.