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

Question

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

My attempt: $$\cos^6 (x) + \sin^4 (x)=1$$ $$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$ $$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$ $$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$

Answer 1

Hint:

Clearly, $\sin x=0\iff\cos^2x=?,$

$\cos x=0\iff\sin^2x=?$ are solutions

For $0< a<1,$ $$a^6<a^4<a^2$$

Answer 2

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 $x \in [0, \frac{\pi}{2}]$.

Answer 3

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 $x = 0, \frac{\pi}{2}$.

Answer 4

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 $\cos x = 0$.

Answer 5

$$1=\cos^6x+\sin^4x\leq\cos^2x+\sin^2x=1,$$ which gives the following system. $$\cos^6x=\cos^2x$$ and $$\sin^4x=\sin^2x.$$ Can you end it now?