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 — using this changing variable $y  = cos^2(x)$ —
$$y^3 + y^2-2y = 0 \Rightarrow y(y^2+y-2) = 0 \Rightarrow y(y+2)(y-1) = 0 $$
As $y = cos^2(x)$ and $0 \leq y \leq 1$:
$$y = 0, 1 \Rightarrow cos^2(x) = 0, 1 \Rightarrow x = 0, \frac{\pi}{2},$$
for $x \in [0, \frac{\pi}{2}]$.
