(c)

Given,

$\sin \left(x+x^2\right)-\sin \left(x^2\right)=\sin x$

$\Rightarrow \quad \sin \left(x+x^2\right)=\sin \left(x^2\right)+\sin x$

$\Rightarrow 2 \sin \left(\frac{x+x^2}{2}\right) \cos \left(\frac{x+x^2}{2}\right)$

$=2 \sin \left(\frac{x^2+x}{2}\right) \cos \left(\frac{x^2-x}{2}\right)$

$\sin \left(\frac{x+x^2}{2}\right)=0$

or $\cos \left(\frac{x^2+x}{2}\right)-\cos \left(\frac{x^2-x}{2}\right)=0$

$\frac{x^2+x}{2}=x \pi$ or $2 \sin \frac{x^2}{2} \sin \frac{x}{2}=0$

$\Rightarrow \frac{x^2+x}{2}=0, \pi, 2 \pi$ or $\frac{x^2}{2}=0, \pi, \frac{n}{2}=0, \pi, 2 \pi$

$\Rightarrow x^2+x=2 \pi$ or $x^2=2 \pi$

$\Rightarrow x^2+x-2 \pi=0$ or $x=\sqrt{2 \pi}$

$x=\frac{-1 \pm \sqrt{1+8 \pi}}{2}$ or $2<\sqrt{2} \pi<3$

$\sqrt{1+8 \pi} \approx \sqrt{25.14}$

$x=\frac{5.2-1-4.2-2.1}{2}$

$\because$ 'Total numbers of solution lies between

$(2,3)=2$