# Solution of trigonometric equation $\sin(n\theta)-2\sin(2\theta)\cos(n\theta)=0$

I have the following equations. \begin{align} & n\theta+\phi=j\pi,\text{ for some }j\in\mathbb{Z},\\ & 2\sin(\theta+\phi)=k(4\cos^2\theta-1)\text{ and }\\ & 2k\cos\theta=\sin\phi. \end{align} I eliminate $k$ and $\phi$ from these equations and get \begin{align}\sin(n\theta)-2\sin(2\theta)\cos(n\theta)=0\end{align} where $n\in\mathbb{N}$. Now I want to find the solution of the above equation.

• I was trying some system of equations. Finally, I have eliminated all other variables and stuck at this point. I am not sure whether an explicit solution to the above equation is possible or not? – G_0_pi_i_e Mar 21 '17 at 4:51
• If you edit the question to show those equations and the place where you got stuck, you might get better quality answers. (No promises.) – David K Mar 21 '17 at 4:57
• $$\sin(n\theta)-2\sin(2\theta)\cos(n\theta)=0\\\to \div \cos(n\theta) \to \\ \tan(n\theta)=2\sin(2\theta)$$ – Khosrotash Mar 21 '17 at 5:03