Solving $\sin(x^2)=\sin(x)^2$ (without transcendence of $\pi$) Is it possible to find all solutions by analytic means? Maybe even without using the fact that $\pi$ does not satisfy a quadratic polynomial with integer coefficients?
 A: We are in a case of relationship $(f \circ g - g \circ f)(x)=0$ with $f = \sin$ and $g:x \rightarrow x^2$. That may or not have in interest. In order to have an idea a general position of roots, we have plotted curves of $f$ and $g$ (we have limited the plot to the first quadrant because both functions are even). There is an infinite number of roots for which a general formula probably does not exist. 

A: First way
Maybe one method could be the usage of Taylor Series. 
Indeed let's try to a low order:
$$\sin x^2 \approx x^2$$
$$\sin^2(x)\approx x^2$$
Which means $$x^2 = x^2$$ not very useful since every real number could be solution of that identity and it won't obviously work. We need more terms.
With more terms, you'll find
$$\sin x^2 \approx x^2 - \frac{x^6}{6}$$
$$\sin^2(x) \approx x^2 - \frac{x^4}{3}$$
thence you have to solve
$$-\frac{x^6}{6} + \frac{x^4}{3} = 0$$
$$-\frac{x^4}{3}\left(\frac{x^2}{2} - 1\right) = 0$$
thence besides the trivial solution $x = 0$ you have $$x^2 = 2$$
More terms more solutions. But surely you won't get them all I guess..
Also this is a numerical method so it's really approximative. But it's a try like everyotherelse 
