JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 29)
The number of solutions of $$\sin ^2 x+\left(2+2 x-x^2\right) \sin x-3(x-1)^2=0$$, where $$-\pi \leq x \leq \pi$$, is ________.
Answer
2
Explanation
$$\begin{aligned} & \sin ^2 x+\left(3-(x-1)^2\right) \sin x-3(x-1)^2=0 \\ & \sin ^2 x+3 \sin x-(x-1)^2 \sin x-3(x-1)^2=0 \\ & \left.\sin x(\sin x+3)-(x-1)^2\right)[\sin x+3]=0 \end{aligned}$$
_5th_April_Evening_Shift_en_29_1.png)
There are two intersections between this graph.
So, Number of solution will be 2 .
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