Solve for sin(x)+sin(2x)+sin(3x)+...+sin(nx)=x The question is $\sin (x)+\sin(2x)+.....+\sin(nx)=x$
where $n$ is any natural number.
I used de Moivre formula to obtain the sum of $\sin(x)+\sin(2x)+...+\sin(nx)$, and I differentiated it to get 
$$
n\sin(x)\sin(nx) \;\;= \;\;3-2\cos(x)+[n\cos(x)-n-1]\cos(nx). 
$$
But I do not know how to continue to solve for $x$.
 A: Equations which mix polynomial and trigonometric terms do not show closed form solutions and only numerical methods would solve the problem.
For the case of the post, first start with the classical identity $$S_n(x)=\sum_{k=1}^n \sin(kx)=\csc \left(\frac{x}{2}\right) \sin \left(\frac{n }{2}x\right) \sin \left(\frac{n+1}{2}
    x\right)$$ the derivative of which being $$S'_n(x)=\sum_{k=1}^n k \cos(kx)=\frac{1}{4} \csc ^2\left(\frac{x}{2}\right) ((n+1) \cos (n x)-n \cos ((n+1) x)-1)$$ So, the equation is $$f(x)=S_n(x)-x=0$$ and, if Newton method is used $$f'(x)=S'_n(x)-1$$ If you look at the plots of $S_n(x)$ and $x$, you will notice that there is no solution for $n=1$ and that for $n>1$ the first root is closer and closer to $0$ when $n$ increases and that the total number of roots is quite limited. Locate one root and use Newton.
For sure, I did not consider the trivial solution $x=0$.
For illustration purposes, let us consider the case where $n=6$ (the smallest value of $n$ for which only one root exists). By inspection, we can see that the solution is close to $\frac{\pi}4$. So, use Newton with $x_0=\frac{\pi}4$. The method will then generate the following iterates : $0.777529$, $0.777656$ which is the solution for six significant figures.
For $n=9$, you will notice three positive roots close to $0.6$, $0.8$ and $1.1$. Let us apply the same method for the middle root. The method will then generate the following iterates : $0.793710$, $0.793697$ which is the solution for six significant figures.
