For all $n \in \Bbb N$ consider the set $A_n = \left \{x \in \Bbb R\ |\ \left (\cos x \right )^n + \left (\sin x \right )^n = 1 \right \}.$ Is $A_n$ uncountanble for all $n \in \Bbb N\ $?
If $n = 2$ then $A_n = \Bbb R$ which is uncountable. How do I prove it for $n \neq 2\ $?
Any help in this regard will be highly appreciated. Thanks in advance.
EDIT $:$ This question is motivated from option $(2)$ of the original question.
On
A very simple argument to handle all integer coefficients $n>2$ is to remember that
$$\sin^2x + \cos^2x=1$$
holds for all $x \in \mathbb R$, from which follows that $\forall x \in \mathbb R: |\sin x| \le 1, |\cos x| \le1$, so for $n > 2$
$$\sin^n x + \cos^n x \le |\sin x|^n + |\cos x|^n \le |\sin x|^2 + |\cos x|^2 = \sin^2x + \cos^2x=1,$$
where the second inequality follows form $\forall t \in [0,1]: t^n -t^2= t^2(t^{n-2}-1) \le 0 \Rightarrow t^n \le t^2$. Equality holds only for $t=0$ or $t=1$, so both $|\sin x|$ and $|\cos x|$ have to be $0$ or $1$, which happens only for a countable set of real $x$ (just either one being $0$ or $1$ is already just a countable set).
$\sin x+\cos x=1$ implies that $1=1^{2}=\sin^{2}(x)+\cos^{2}(x)+2\sin (x) \cos (x)$ which gives $\sin (2x)=0$. Hence $2x$ is a an integer multiple of $\pi$. Hence $A_1$ is countable.
Using Complex Analysis it is easy to see that $A_n$ is uncoutable only for $n=2$. If $A_n$ is uncountable it has a limit point. A Theorem in Complex Analysis shows that we must have $\cos^{n} x+\sin^{n}x=1$ for all $x$. Put $x=\frac {\pi} 4$ to see that $n=2$.
Real analytic proof for $n$ even: On $A_n$ we have $t^{m}+(1-t)^{m}=1$ wheer $m=\frac n 2$ and $t=\sin ^{2}x$. This equation is a polynomial equation in $t$ and it has only finitely many roots. Hence there are only finite number of possible values for $\sin ^{2}x$ and each of then give only countably many values of $x$.