I want to solve the following equation but I don't know the right way, I've tried some things such as converting sines to their exponentials through euler's equation but still I can't go on. Any ideas? Here is the equation: $$\sin (b-u) + \sin u \ e ^{ \frac{-b}{0.333} } = 0 $$
Where $u$ is known variable and is equal to $20$ degrees. The variable that has to be found is $b$ .
Working with radians and whole numbers, you are looking for the zero's of function $$f(b)=\sin \left(b-\frac{\pi }{9}\right)+e^{-3 b} \sin \left(\frac{\pi }{9}\right)$$ which, being transcendental, will not show analytical solution.
I suppose that you already noticed the trivial solution $b=0$. But, graphing, we notice another one close to $0.05$.
Using Newton method, you would get the following iterates $$\left( \begin{array}{cc} n & b_n \\ 0 & 0.050000000 \\ 1 & 0.053424485 \\ 2 & 0.053215971 \\ 3 & 0.053215198 \\ 4 & 0.053215198 \end{array} \right)$$
Let us be even more lazy : the root being small, use Taylor expansion to get $$f(b)=b \left(\cos \left(\frac{\pi }{9}\right)-3 \sin \left(\frac{\pi }{9}\right)\right)+5 b^2 \sin \left(\frac{\pi }{9}\right)+O\left(b^3\right)$$ which gives, as an estimate, $$b=\frac{1}{5} \left(3-\cot \left(\frac{\pi }{9}\right)\right)\approx 0.0505045$$
If we add the next term of the expansion $$f(b)=b \left(\cos \left(\frac{\pi }{9}\right)-3 \sin \left(\frac{\pi }{9}\right)\right)+5 b^2 \sin \left(\frac{\pi }{9}\right)-b^3 \left(\frac{9}{2} \sin \left(\frac{\pi }{9}\right)+\frac{1}{6} \cos \left(\frac{\pi }{9}\right)\right)+O\left(b^4\right)$$ which is a quadratic in $b$. Solving it, the estimate would be $0.053324$.
To polish the root, one iteration of Newton method would probably suffice.