Solve $\cos^n x + \sin^n x =1 $ the solutions of this equation as a function of the value of $n$??         
\begin{align}      
 \cos^n x + \sin^n x =1 
\end{align}
I already found the solution if n is odd, 
 A: Hint: 
For positive $n$: You know that for any $x$, $$\cos^2 x + \sin^2 x = 1,$$ and both the terms on the left are positive. For $n = 2k$, if $x$ is a solution, then you have $\cos^{2k} x + \sin^{2k} x = 1$, so $$(\cos^2 x)^k + (\sin^2 x)^k = 1$$. 
Look at those two displayed equations and ask yourself, "How are $\cos^2 x $ and $(\cos^2 x)^k$ related? Which is larger in general?"
For negative $n$: A similar argument should work, but with the inequality reversed. 
A: If $n$ is odd, we can write it as:$n = 2k+1, k \ge 1$, $1 = |\sin^{2k+1}x + \cos^{2k+1}x| \le |\sin x|\cdot \sin^{2k}x + |\cos x|\cdot \cos^{2k}x \le \sin^{2k}x+\cos^{2k}x \le \sin^2 x + \cos^2x = 1$. thus $|\sin x| = 1, 0$ , and you can find $x$ from this. If $n = 1$, then $\sin x + \cos x = 1\implies (\sin x+\cos x)^2 = 1 \implies \sin (2x) = 0 \implies 2x = m\pi \implies x = \dfrac{m\pi}{2}, m \in \mathbb{Z}$. If $n$ is even, then $n \ge 2 \implies 1 = \sin^2x + \cos^2 x \ge \sin^n x+\cos^n x = 1\implies \cos^2 x = 1, 0 \implies \cos x = 0, \pm 1 \implies x = m\pi, \pm\dfrac{\pi}{2} + 2m\pi, m \in \mathbb{Z}$
A: There is a nice geometrical interpretation if you will.
We know $\cos^2 x + \sin^2 x = 1$ for all $x$. You want to solve $\cos^n x + \sin^n x = 1$. If $a = \cos x$ and $b = \sin x$, then you want to solve the simultaneous equations:
$$
a^2 + b^2 = 1,\quad\quad\mbox{circunference}\\
a^n + b^n = 1,\quad\mbox{super-circunference}
$$
The circunference and super-circunference for $n > 2$ will intercept only in the points $(1, 0)$, $(0, 1)$, $(-1, 0)$, $(0, -1)$. The greater than $n$, the more the super-circunference will look like a square.
A: If $\cos x\sin x\not=0$, then $|\cos x|=\sqrt{1-\sin^2x}$ and  $|\sin x|=\sqrt{1-\cos^2x}$ are both strictly less than $1$, which implies $\cos^nx+\sin^nx\lt\cos^2x+\sin^2x=1$ for $n\gt2$.  For $n=1$, $\cos x+\sin x=1$ implies $\cos^2x+2\cos x\sin x+\sin^2x=1$, which implies $\cos x\sin x=0$.  In sum, when $n$ is a positive integer other than $2$, $\cos^nx+\sin^nx=1$ implies $x$ is a multiple of $\pi/2$.  If $n$ is even, any (integer) multiple of $\pi/2$ is a solution.  If $n$ is odd, the multiple must be congruent to $0$ or $1$ mod $4$.
For negative integers $n$, we cannot have $\cos x\sin x=0$.  If $n\lt0$ is even, then $\cos x\sin x\not=0$ implies $\cos^n+\sin^n\gt\cos^2+\sin^2=1$, so the equation has no solutions.  If $n\lt0$ is odd, however, there are solutions.  In particular, there is always a solution with $-\pi/2\lt x\lt 0$, since $\sec x\to\infty$ as $x\to-\pi/2^+$ while $\csc x\to-\infty$ as $x\to0^-$.
Finally, if $n=0$, the equation presumably has no solutions, since $\cos^0x+\sin^0x=1+1=2\not=1$ if $\cos x\sin x\not=0$ and isn't clearly defined if $\cos x\sin x=0$.
