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:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
 A: 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 $x \in [0, \frac{\pi}{2}]$.
A: 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 $x = 0, \frac{\pi}{2}$.
A: Hint:
Clearly, $\sin x=0\iff\cos^2x=?,$
$\cos x=0\iff\sin^2x=?$ are solutions
For $0< a<1,$ $$a^6<a^4<a^2$$
A: 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 $\cos x = 0$.
A: $$1=\cos^6x+\sin^4x\leq\cos^2x+\sin^2x=1,$$ which gives the following system.
$$\cos^6x=\cos^2x$$ and $$\sin^4x=\sin^2x.$$
Can you end it now?
