<s>[INST] Finding All Solutions For $\sin(x) = x^2$ Hello everyone how can I find the count of the solution for $\sin(x) = x^2$?
I know there is a one solution in $x = 0$ and for the other solutions I tried to find the extreme point of the function: $y = x^2 - \sin(x)$ and $y'$ is:
$y' = 2x -\cos(x)$ but I don't know how to solve this equation.
 [/INST] Just for the fun of it !
There is no explicit solution for the zero of function $$f(x)=2x -\cos(x)=0$$ If you need it, use Newton method which will converge quite fast as shown in the table
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.000000 \\
 1 & 0.500000 \\
 2 & 0.450627 \\
 3 & 0.450184
\end{array}
\right)$$ Another solution could be a series expansion
$$2x -\cos(x)=1-2 x-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ and use series reversion to get
$$x=t-\frac{t^2}{4}+\frac{t^3}{8}-\frac{11 t^4}{192}+\frac{3 t^5}{128}-\frac{121
   t^6}{23040}-\frac{19 t^7}{5120}+O\left(t^8\right)\quad \text{where}\quad t=\frac{1-f(x)}2$$ Making $f(x)=0$ that is to say $t=\frac 12$, you should get, as an approximation,
$$x =\frac{531037}{1179648}\approx 0.450166$$
Amazing would be to use the $\color{red}{1,400}$ years old approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$  which would lead to the cubic equation
$$2 x^3+4 x^2+2 \pi ^2 x-\pi ^2=0$$
$$x=-\frac{2}{3} \left(1-\sqrt{3 \pi ^2-4} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{63 \pi ^2-32}{4 \left(3 \pi
   ^2-4\right)^{3/2}}\right)\right)\right)\approx 0.449785$$
</s>