# Solving a linear trigonometric equation

Let $n$ be a natural number. For $a_i,\omega_i,\varphi_i \in \mathbb{R}$ how can one find solutions $x \in \mathbb{R}$ for the equation:

$$\sum_{i=1}^n a_i \cos( \omega_i \cdot (x-\varphi_i)) = 0$$

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I feel exact solution is likely to be hard/impossible, if the $\omega_i$'s are incommensurable (as in the example, $\omega_1 = 1$, $\omega_2 = \sqrt{2}$ and $\omega_3 = e+\pi$). – Srivatsan Sep 1 '11 at 11:25
Analytically? You're outta luck; a transcendental equation like that ain't that easy. Numerically? Well, sure... Newton-Raphson's pretty darn good, as long as you have nice starting estimates. – J. M. Sep 1 '11 at 11:27
For $\sum_{i=1}^{2}a_{i}\cos (\omega _{i}(x-\varphi _{i}))=0$, SWP gives the following response: solution is $x=\frac{\rho _{1}+\omega _{2}\varphi _{2}}{ \omega _{2}}$, where $\rho _{1}$ is a root of $\hat{Z}\omega _{1}+\omega _{1}\omega _{2}\varphi _{2}-\omega _{1}\varphi _{1}\omega _{2}+\omega _{2}\pi -\omega _{2}\arccos a_{2}\frac{\cos \hat{Z}}{a_{1}}=0$. – Américo Tavares Sep 1 '11 at 11:41
@Américo: that looks positively scary. I'd be trying for NR instead if I were Alex. – J. M. Sep 1 '11 at 11:57