<s>[INST] 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, 
 [/INST] 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 $n \ge 2 \implies 1 = \sin^2x + \cos^2 x \ge \sin^n x+\cos^n x = 1\implies \cos^2 x = 1, 0 \implies \cos x = 0, \pm 1 \implies x = m\pi, \pm\dfrac{\pi}{2} + 2m\pi, m \in \mathbb{Z}$
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