This question is based on the question I posted before. The link is Maxima and minima of sinusoidal function of harmonics
In need to find the root of following equation $$\ \omega \cdot \cos(5\omega t)+\ \omega\cdot \cos(7\omega t)+\ \omega\cdot \cos(9 \omega t)+\ \omega\cdot \cos(11 \omega t)+\ \omega\cdot \cos(13 \omega t) = 0$$
In Matlab Chebyshev polynomial function is $ f(x) = ChebyshevT(n,x) $ giving you the $n-$th root. If I substitue $n = 5\cdot \omega,~ 7 \cdot \omega~ $ etc then the Chebyshev polynomial becomes too complicated since $\omega = 2 \cdot \pi \cdot f$.
My way of doing it was replace $\omega \cdot t = \theta$. So the above equation changes to $$\ \frac \theta t \cdot ( \cos(5\theta)+\ \cos(7\theta)+\ \cos(9 \theta)+\cos(11\theta)+ \cos(13 \theta)) = 0$$ and then find the roots of function inside the brackets. Then $\theta $ can be re-substituted to find the value of $t$.
I used Matlab to find the roots using the equation with cosines, and then replacing it with Chebyshev Polynomials of first kind. However the results are not at all even close. Also i am bit confused about what is to be done with $\frac \theta t$ outside the brackets. Any help will be greatly appreciated.
Considering that the equation is the real part of a sum of imaginary exponentials,
$$e^{5i\alpha}(1+e^{2i\alpha}+e^{4i\alpha}+e^{6i\alpha}+e^{8i\alpha})=e^{5i\alpha}\frac{e^{10i\alpha}-1}{e^{2i\alpha}-1}=e^{5i\alpha}\frac{\sin5\alpha e^{5i\alpha}}{\sin\alpha e^{i\alpha}}=\frac{\sin5\alpha}{\sin\alpha} e^{9i\alpha}.$$
The real part cancels when
$\sin5\alpha$ is zero but $\sin\alpha$ is not, or
$\cos9\alpha$ is zero.
We used
$$e^{i\alpha}-1=\cos\alpha-1+i\sin\alpha=2i\sin\frac\alpha2e^{i\alpha/2}.$$