If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$

Hint:
Clearly, $\sin x=0\iff\cos^2x=?,$
$\cos x=0\iff\sin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
