Proof for $\frac{\cos^6 x}{\cos 6x}=1$ 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.
 A: $$x=\frac{\pi}{2}$$
$$\frac{\cos^6{(x)}}{\cos{(6x)}}=0$$
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
$$31y^3-48y^2+18y-1=0$$
$$(y-1)(31y^2-17y+1)=0$$
$$y=1,\frac{17\pm\sqrt{165}}{62}$$
$$\therefore\cos^2{(x)}=1,\frac{17\pm\sqrt{165}}{62}$$
$$x=\pm\pi, \pm0.8040277545, \pm1.308946784, \pm 1.83264587,\pm2.337564899, ...$$
A: 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 $k\in\mathbb Z$.


*$t=\frac{17+\sqrt{17^2-4\cdot31}}{62}=\frac{17+\sqrt{165}}{62}.$
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
$$x=\pm\frac{1}{2}\arccos\frac{-14-\sqrt{165}}{31}+\pi k.$$
