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, 

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 $\cos x\sin x=0$.
