Methodology to solve trigonometric equations such as $4x+\tan(x)=\pi$

Since it is transcendental, there is no explicit expression for the zero of function $$f(x)=4x+\tan(x)-\pi$$ and numerical methods (or approximations) are required.

Being very lazy, expand $f(x)$ as a series around $x=0$ and obtain $$f(x)=-\pi +5 x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+\frac{62 x^9}{2835}+\frac{1382 x^{11}}{155925}+O\left(x^{13}\right)$$

Now, using series reversion, $$x=t-\frac{t^3}{15}-\frac{t^5}{75}-\frac{t^7}{7875}+\frac{67 t^9}{70875}+\frac{5017 t^{11}}{19490625}+O\left(t^{13}\right)$$ where $t=\frac{f(x)+\pi}{5}$. Since we desire $f(x)=0$ then $t=\frac \pi 5$. Replacing, the estimate is then $$x \sim \color{red}{0.610487}235$$ while the exact solution given by Newton method is $0.610487119$

Just for the fun of doing it almost by hand, all of the above was done using my $50$ years old non-programmable calculator.