How to solve $\tan x=x^2$ in radians? How to solve $\tan x=x^2$ with $x \in [0, 2\pi]$? I try with trigonometry and many ways but the numerical solutions seems to be difficult.

Solving $\tan(x)=x^2$ is the same as finding the intersection of the two curves $y=\tan(x)$ and $y=x^2$.
If you plot these two curves for $x \in [0, 2\pi]$, you will notice that, beside the trivial root $x=0$, they intersect just before $x=\frac{3\pi}{2}$. The function being very stiff, it is then better to search for the solution of $\sin(x)=x^2\cos(x)$ as already mentioned by Jack D'Aurizio. 
For getting the solution, there are different ways. First, to get an approximation, you can expand the expression as a Taylor series built at $x=\frac{3\pi}{2}$. This gives, for the second order, $$\sin(x)-x^2\cos(x)=-1-\frac{9}{4} \pi ^2 \left(x-\frac{3 \pi }{2}\right)+\left(\frac{1}{2}-3 \pi \right)
   \left(x-\frac{3 \pi }{2}\right)^2+O\left(\left(x-\frac{3 \pi }{2}\right)^3\right)$$ and, solving the quadratic, the solution to be retained is $$x = \frac{-12 \pi +54 \pi ^2+\sqrt{128-768 \pi +324 \pi ^4}}{48 \pi -8}\simeq 4.66651$$ which is not bad since the exact solution is $4.66650$ as already mentioned by Jack D'Aurizio. 
To reach accurate solutions, Newton method, starting with a guess $x_0=\frac{3\pi}{2}$ will updtate it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ and the iterates will be $4.667357343$, $4.666499882$, $4.666499563$ which is the solution for ten significant figure.
We could have replace Newton by Halley or Housholder methods; they require higher derivatives but less iterations. I do not think that it could be of major interest for this specific problem.
Just for illustration, Halley method gives as first iterate $$x_1=\frac{3 \pi }{2}-\frac{4 \left(-8+48 \pi +81 \pi ^4\right)}{729 \pi ^6} \simeq 4.666542356$$ while Householder would give $$x_1=\frac{3 \pi }{2}-\frac{4 \left(128+3 \pi  \left(3 \pi  \left(448-48 \pi ^2+432 \pi
   ^3+729 \pi ^6\right)-512\right)\right)}{59049 \pi ^{10}} \simeq 4.666501749$$
