Minima and maxima of $\left({\frac{\sin 10x}{\sin x}}\right)^2$ on $[0,\pi]$ 
Find the minima and maxima of $\left({\frac{\sin 10x}{\sin x}}\right)^2$in the interval $\left [ 0,\pi \right ]$.

This is a question from BdMO that still haunts me a lot. I would like to find an answer to this question.
 A: The function is positive, so that the minima are the points where the function equals $0$, that is, $k\,\pi/10$, $1\le k\le 9$. Between each two conaecutive zeros, the functions has a local maximum. The function is symmetric around $\pi/2$; it is enough to study $[0,\pi/2]$. It is easy to see that the function is decreasing on $[0,\pi/10]$. The maximum in this interval is
$$
\lim_{x\to0}\Bigl(\frac{\sin 10\,x}{\sin x}\Bigr)^2=100.
$$
If $k\,\pi/10\le x\le(k+1)\,\pi/10$, $1\le k\le4$ then
$$
\Bigl(\frac{\sin 10\,x}{\sin x}\Bigr)^2\le\frac{1}{\sin^2(k\,\pi/10)}\le\frac{100}{k^2\,\pi^2}<100.
$$
The maximum is attained at $x=0$ (and $x=\pi$).
A: If you consider function $$f(x)=\left({\frac{\sin 10x}{\sin x}}\right)^2$$, its derivative is given by $$f'(x)=-2 \sin (10 x) \csc ^2(x) \Big(\sin (10 x) \cot (x)-10 \cos (10 x)\Big)$$ so the extrema are given by the solutions of $$\sin(10x)=0$$ and the solutions of $$\sin (10 x) \cot (x)-10 \cos (10 x)=0$$ The first one does not make any problem, the second one is harder (it has ten solutions which are closer and closer to the values where tangents are undefined).
Concerning the minimum of $f(x)$, it is clearly $0$ nine times and the largest maximum it is $100$ obtained at the limits of the interval.
A: Let's say we name the function $f(x)$. Just do the derivative $f^\prime(x)$ and find $x$ such that $f^\prime(x) = 0$. If I did not make a mistake, you should get that $f^\prime(x) = 0$ if any of these 2 conditions hold:


*

*$\sin(10x) = 0$

*$10 \tan(x) = \tan(10x)$


The second derivative tells you whether these values are minima or maxima. The results obtained from the second equation always have negative second derivative, so they correspond to maxima.
Notice that the function is not defined at $x=0$. However, if you want to make it continuous at that point, notice that its value should be defined as $f(0)=100$. This is one maximum.
A: Since no one has solved the hard problem of solving:
$$\tan(10x)=10\tan x \tag1$$
Which gives the location of the rest of the extrema points, I'll give it a shot.
Using the identity:
$$\tan nx = \frac{\tan(nx-x)+\tan x}{1-\tan(nx-x)\tan x}$$
One can prove by induction that for even $n$:
$$\tan nx = \frac{\displaystyle\sum_{i=1}^{n/2} (-1)^{i+1} \binom {n} {2i-1}\tan^{2i-1} x}
{\displaystyle\sum_{i=0}^{n/2} (-1)^{i} \binom {n} {2i}\tan^{2i} x}$$
This leads to equation $(1)$ becoming a $10$th degree polynomial in $\tan x$. Actually writing out the coefficients, leads in this particular case to a quartic equation in $s=t^2$, and the additional solution $t=0$:
$$5s^4-220 t^3+990s^2-924 s-164=0 \tag2$$
Thus the $9$ solutions can actually be expressed as an $\arctan$ of explicit algebraic numbers $s_n$ which are the solutions of $(2)$:
$$x=\arctan{\sqrt{s_n}}$$ 
