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

Question

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

Professor PC Thomas Classes · Answer

(2): We have, $$ \begin{aligned} \sin ^{2} x+\left(2+2 x-x^{2}ight) \sin x-3(x-1)^{2} & =0 \ 	ext { where }-\pi & \leq x \leq \pi \end{aligned} $$ $\Rightarrow \quad \sin ^{2} x+\left(3-(x-1)^{2}ight) \sin x-3(x-1)^{2}=0$\ $\Rightarrow \quad \sin x(\sin x+3)-(x-1)^{2}(\sin x+3)=0$\ $\Rightarrow \quad(\sin x+3)\left(\sin x-(x-1)^{2}ight)=0$\ $\Rightarrow \quad \sin x+3 
eq 0$\ $\Rightarrow \quad \sin x-(x-1)^{2}=0$\ $\Rightarrow \quad \sin x=(x-1)^{2}$\ There are two points of intersection.\ So, number of solutions are 2 .\