<s>[INST] 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:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
 [/INST] Let $\cos x = c, \sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $\sin x = 0$ or $\cos x = 0$.
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