I have to solve the following trigonometric equation:
$$ 4\cdot \arctan(\omega_{180}) \ + \ \omega_{180} \ = \ \pi $$ which by setting $\arctan(\omega_{180}) = x$ becomes:
$$ 4x \ + \ \tan(x) \ - \ \pi \ = \ 0 $$
In order to solve the latter equation in terms of $x$, I tried to replace the term of $\tan(x)$ by the Taylor series approximation of $\sin(x), \cos(x)$ using only the first two terms but the derived results were not correct. Anyway, I used MATLAB to solve it and it did solved it. The correct result is:
$$ x = 0.61048 \Rightarrow \omega_{180} = 0.7 \text{rad/sec} $$
What I would like to ask is in which ways can I solve this equation by hand ? I don't want a complete solution as an answer but rather the methodologies I could study and then apply in such trigonometric equations.
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.