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,

Solution 1:

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.

Solution 2:

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 $n \ge 2 \implies 1 = \sin^2x + \cos^2 x \ge \sin^n x+\cos^n x = 1\implies \cos^2 x = 1, 0 \implies \cos x = 0, \pm 1 \implies x = m\pi, \pm\dfrac{\pi}{2} + 2m\pi, m \in \mathbb{Z}$

Solution 3:

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.