I need x form "x tanx=0.04"

This may be difficult to solve algebraically. Try Newton's method.

x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

So, we have:

x-\frac{xtan(x)-.04}{tan(x)+xsec^{2}(x)}

There are several solutions.

Let the initial guess be {\pi}.

It should converge quickly to one such solution.