The number of solutions of where is, JEE Main Previous Year Trigonometry Questions and Solutions - Solution of Triangles-2024-12

Question

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 _________. [2024]

Smart Achievers · Accepted Answer

(2)

We have $\sin^{2} x + (2 + 2 x - x^{2}) \sin x - 3 {(x - 1)}^{2} = 0$ where $- π \leq x \leq π$

$\Rightarrow \sin^{2} x + (3 - {(x - 1)}^{2}) \sin x - 3 {(x - 1)}^{2} = 0$

$\Rightarrow \sin x (\sin x + 3) - {(x - 1)}^{2} (\sin x + 3) = 0$

$\Rightarrow (\sin x + 3) (\sin x - {(x - 1)}^{2}) = 0$

$\Rightarrow \sin x + 3 \neq 0 \Rightarrow \sin x - {(x - 1)}^{2} = 0 \Rightarrow \sin x = {(x - 1)}^{2}$

There are two points of intersection.

So, number of solutions are 2.

Want Us to Email you About Special Offers & Updates?