Solve trig function for $x$

Question

For what value(s) of $x$ does the following function satisfy?

$$ \dfrac{-16}{9} = \dfrac{\cos(18x)}{\cos (24x)} $$

Unsure if I need an identity to solve or it's just basic.

Answer 1

This is not a nice problem.

As mentioned in the comments, let $y=6x$. Then we have $$-\frac{16}9=\frac{\cos(3y)}{\cos(4y)}\implies 16\cos(4y)=-9\cos(3y)\tag{1}$$ so $$16(8\cos^4y-8\cos^2y+1)=-9(4\cos^3y-3\cos y)$$ giving $$128\cos^4y+36\cos^3y-128\cos^2y-27\cos y+16=0$$

Hence we solve $$128t^4+36t^3-128t^2-27t+16=0$$ with $t=\cos y$, which is a quartic with non-integer real solutions.

We use the Ferrari-Cardano Method. After some tedious calculations we find that $$\Delta=2.258760084...\times10^{12}$$$$\Delta_1=20866688$$$$\Delta_0=43876$$$$p=-\frac{8435}{8192}$$$$q=-\frac{1132992}{16777216}$$ so $$Q=\sqrt[3]{\frac{\Delta_1+\sqrt{-27\Delta}}{2}}=\sqrt[3]{10433344+3904693.402...i}$$$$S=\frac12\sqrt{-\frac23p+\frac1{3a}\left(Q+\frac{\Delta_0}Q\right)}=\frac12\sqrt{\frac{8435}{12288}+\frac1{384}\left(Q+\frac{43876}Q\right)}$$ and finally the solutions are $$x_{1,2}=-\frac b{4a}-S\pm\frac12\sqrt{-4S^2-2p+\frac qS}$$ and $$x_{3,4}=-\frac b{4a}+S\pm\frac12\sqrt{-4S^2-2p+\frac qS}$$

In the end, the solutions are $$t=0.908425,0.279770,-0.515677,-0.963768$$

Therefore, $$x=\frac y6=\frac{\cos^{-1}t}6$$ would give all the solutions to $(1)$.