Solve $\cos^{n}x-\sin^{n}x=1$ with $n\in \mathbb{N}$. Solve $\cos^{n}x-\sin^{n}x=1$ with $n\in \mathbb{N}$
I have no idea how to deal with this crazy question. One idea came into my mine is factorization, but I can't go on... Can anyone help me please? Thank you.
 A: We consider the $2\pi$-periodic function
$$f(x):=\cos^n x-\sin^n x$$
and determine its stationary points in $[0,2\pi[\ $. One gets
$$f'(x)=-n\cos x\sin x\bigl(\cos^{n-2}x+\sin^{n-2}x\bigr)\ ;$$
therefore the stationary points  are the multiples of ${\pi\over2}$, and for odd $n>2$ the points where $\cos x=-\sin x$, i.e., the points ${3\pi\over4}$ and ${7\pi\over4}$. In these points one has the values
$$f(0)=1, \quad f({\pi\over2})=-1,\quad f(\pi)=(-1)^n,\quad f({3\pi\over2})=(-1)^{n-1}\ ,$$
furthermore for  $n=2m+1$ the values
$$f({3\pi\over4})=-{\sqrt{2}\over 2^m}, \quad f({7\pi\over4})={\sqrt{2}\over 2^m}<1\ .$$
It follows that the global maximal value of $f$ is $1$. This value is taken at $0$ and $\pi$ if $n$ is even, and at $0$ and ${3\pi\over2}$ if $n$ is odd.
A: Hint:
For all $n$, when $\cos(x)=1$, $\sin(x)=0$.
For even $n$, when $\cos(x)=-1$, $\sin(x)=0$.
For odd $n$, when $\sin(x)=-1$, $\cos(x)=0$.
A: If $n$ is even, then 
$$1= \cos^{n}x-\sin^{n}x \leq 1-0=1$$
with equality if and only if $\cos^{n}x=1, \sin^n(x)=0$.
If $n$ is odd,
$$1= \cos^{n}x-\sin^{n}x \,,$$
implies $\cos(x) \geq 0$ and $\sin(x) <0$. Let $\cos(x)=y, \sin(x)=-z$, with $y,z \geq 0$.
$$y^n+z^n=1$$
$$y^2+z^2=1$$
Case 1: $n=1$:
Then , since $0 \leq y,z \leq 1$ we have 
$$1 =y+z \geq y^2+z^2 =1$$
with equality if and only if $y=y^2, z=z^2$.
Case 2: $n \geq 3$:
Then , since $0 \leq y,z \leq 1$ we have 
$$1 =y^2+z^2 \geq y^n+z^n =1$$
with equality if and only if $y^2=y^n, z^2=z^n$.
