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 If we rewrite $u = \cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $\cos^6 x \leq \cos^2 x$ and $\sin^4 x \leq \sin^2 x$, and in both cases equality holds (for $x \in [0, \frac{\pi}{2}]$) only for $x = 0, \frac{\pi}{2}$.
