<s>[INST] How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion. I replaced 1 by $\cos^2(x) + \sin^2(x) $ hoping to factor it but to no avail.
 [/INST] Write $u = \cos x$, $v=\sin x$. From $u^2 + v^2=1$ we get  $u$, $v$ $\le 1$  so $u^3\le u^2$, $v^3 \le v^2$, and adding up we get $u^3 + v^3\le 1$. We have equality if and only if we have equality in the previous equalities, therefore $(u,v)=(1,0)$ or $(0,1)$.
It is interesting to sketch the curves $u^2+v^2=1$ and $u^3+v^3=1$. From the above, the function $u^3+v^3$ has maximum value $1$ on $u^2+v^2=1$, so the two curves are tangent at $(1,0)$ and $(0,1)$. In other words, the multiplicities of the intersection at $(1,0)$, $(0,1)$ are $2$. Bézout's theorem tells us there should be two more intersection points. They are complex,  $(-1+\frac{i}{\sqrt{2}},-1-\frac{i}{\sqrt{2}})$ and its conjugate $(-1-\frac{i}{\sqrt{2}},-1+\frac{i}{\sqrt{2}})$

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