Proof for $\frac{\cos^6 x}{\cos 6x}=1$

Question

I tried some transformations but nothing is bringing me to 1.

$$\frac{\cos^6 x}{\cos 6x} = 1$$

I tried something like $\left(\cos^2 x\right)^3$, then $\left(2\cos^2x-1\right)^3$. And, for $\cos 6x$, to transform into $\cos(2(3x))$ or $\cos(3 (2x))$. But it's not going well.

Answer 1

$$x=\frac{\pi}{2}$$ $$\frac{\cos^6{(x)}}{\cos{(6x)}}=0$$ Otherwise for specific values we need $$\cos^6{(x)}=\cos{(6x)}$$ $$=\Re{(\cos{(x)}+i\sin{(x)})^6}$$ $$=32\cos^6{(x)}-48\cos^4{(x)}+18\cos^2{(x)}-1$$ So by letting $y=\cos^2{(x)}$ we have $$31y^3-48y^2+18y-1=0$$ $$(y-1)(31y^2-17y+1)=0$$ $$y=1,\frac{17\pm\sqrt{165}}{62}$$ $$\therefore\cos^2{(x)}=1,\frac{17\pm\sqrt{165}}{62}$$ $$x=\pm\pi, \pm0.8040277545, \pm1.308946784, \pm 1.83264587,\pm2.337564899, ...$$

Answer 2

Let $\cos^2x=t$.

Thus, since $$\cos3\alpha=4\cos^2\alpha-3\cos\alpha,$$ we obtain: $$\cos^6x=\cos6x$$ or $$t^3=2(4\cos^3x-3\cos x)^2-1$$ or $$t^3=2t(4t-3)^2-1$$ or $$31t^3-48t^2+18t-1=0$$ or $$31t^3-31t^2-17t^2+17t+t-1=0$$ or $$(t-1)(31t^2-17t+1)=0.$$ 1. $t=1$.

Thus, $$\cos^2x=1$$ or $$\sin{x}=0$$ or $$x=\pi k,$$ where $k\in\mathbb Z$.

2. $t=\frac{17+\sqrt{17^2-4\cdot31}}{62}=\frac{17+\sqrt{165}}{62}.$

Here, $$\frac{1+\cos2x}{2}=\frac{17+\sqrt{165}}{62}$$ or $$\cos2x=\frac{\sqrt{165}-14}{31},$$ which gives $$x=\pm\frac{1}{2}\arccos\frac{\sqrt{165}-14}{31}+\pi k.$$ 3. $t=\frac{17-\sqrt{165}}{62}.$

Here, $$\cos2x=\frac{-14-\sqrt{165}}{31},$$ which gives $$x=\pm\frac{1}{2}\arccos\frac{-14-\sqrt{165}}{31}+\pi k.$$