Trying to solve for x when you have sine and cosine in the function Graph $f(x) = \sin (x^2)$
I need to graph $f(x) = \sin (x^2)$ from $-2\pi$ to $2\pi$ and from that, I need to include the first derivative which is set to zero and used to find the maximum and minimum.
That I totally get however then I need to go and find the second derivative and set equal to zero and find any points of inflection. (Basically, find 2nd derivative, set equal to zero, and find $x$)
However the second derivative I found is $2\cos(x^2) - 4x^2 \sin(x^2)$ and because of the cosine and sine, I have no idea how to solve for $x$.
Any help will be deeply appreciated. Just to be clear, I'm having problems trying to solve for $x$ in the second derivative.
 A: We can do a few simplifications: Substitute $z=x^2$ so $x = \pm\sqrt z$. Then
$$\begin{align*}
2\cos x^2 - 4x^2 \sin x^2 & = 0\\
\Leftrightarrow \cos z - 2z\sin z & = 0 & z \ge 0
\end{align*} $$
Now since $z=0$ is no solution and $\cos z = 0 \Rightarrow \sin z \ne 0$ is also no solution, we can divide by $\cos z$ and get
$$1 = 2z\tan z$$
so
$$z = \frac12 \cot z$$
The solutions for the latter seem symbolically intangible, but we can use the numerical results to obtain numerical results for $x = \pm\sqrt z$
A: Just a sketch for $x>0$: this transcendental equation cannot be solved explicitly. Nevertheless, by putting it in the form $\tan x^2 = \frac 1 {2x^2}$ one notices that the graph of the left-hand side (LHS) is made of several disjoint branches, each one increasing continuously from $0$ to $\infty$, while the right-hand side (RHS) is decreasing from $\infty$ to $\frac 1 {8 \pi^2}$. As such, the LHS will intersect the RHS in as many points as branches the LHS has and these will be the inflection points. A similar analysis for $x<0$.
