The problem is:

$\textrm{From the following equation}$


$\textrm{find the value of:}$

$$\sec{\omega}+\sec{3\omega}$$ $\textrm{Consider:}$ $$\omega\neq (2K+1)\frac{\pi}{6}\;k\in \mathbb{Z}$$

I figured out some familiar expression in the third term of the equation and by applying prosthaphaeresis formula in the third term of the equation I got to this,



However there are second and double angles in the latter equation, therefore I decided to transform into their power equivalents shown below,


By using pitagorean identity in the earlier equation then I reached to this expression:




to which is transformed into a cubic equation as follows (note I multiplied by $-1$):


I am not sure if my procedure is correct, moreover to solve a cubic equation entitles a problem since I don't know how to find the angle from there. How can I get to the answer?


Consider the equation $$11(2\cos^{2}\omega-1)-14(4\cos^{3}\omega-3\cos\omega)=0$$ and just multiply it by $2$: $$11(4\cos^{2}\omega-2)-28(4\cos^{3}\omega-3\cos\omega)=0$$ From this equation and the assumption that $\omega \neq (2k+1)\frac{\pi}{6}$, we conclude that $$\frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\omega}=\frac{28}{11}$$ But it is easy to check $ \frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\omega} = \sec \omega + \sec 3\omega $.

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    $\begingroup$ Thanks for that, now by comparing with your reasoning I figured out that I was missing. It makes me glad to know I was not that far off from the answer. $\endgroup$ – Chris Steinbeck Bell Nov 4 '17 at 0:25

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