# How to solve a trigonometric equation involving double and triple angles?

The problem is:

$\textrm{From the following equation}$

$$10\cos{2\omega}-13\cos{3\omega}+2\sin{\frac{5\omega}{2}}\sin{\frac{\omega}{2}}=0$$

$\textrm{find the value of:}$

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

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,

$10\cos{2\omega}-13\cos{3\omega}-(\cos3\omega-cos2\omega)=0$

$11\cos{2\omega}-14\cos{3\omega}=0$,

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

$11(\cos^{2}\omega-\sin^{2}\omega)-14(4\cos^{3}\omega-3cos\omega)=0$,

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

$11(\cos^{2}\omega-(1-\cos^2\omega))-14(4\cos^{3}\omega-3cos\omega)=0$,

$11(2\cos^{2}\omega-1)-14(4\cos^{3}\omega-3cos\omega)=0$,

$22\cos^{2}\omega-22-56\cos^{3}\omega+42\cos\omega=0$

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

$56\cos^{3}\omega-22\cos^{2}\omega-42\cos\omega+22=0$

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 $\frac{4\cos^{2}\omega-2}{4\cos^{3}\omega-3\cos\omega} = \sec \omega + \sec 3\omega$.