Solve the equation $\cos^n(x) - \sin^n(x)=1$ Solve the equation $\cos^n(x) - \sin^n(x)=1,n \in \mathbb{N}-\{0\}$




If $n$ is even then $\cos^n(x) = \sin^n(x)+1$ is only possible if $\sin(x)=0$ therefore the solution is $x=k\pi, k \in \mathbb{Z}$.

I'm having problems with $n$ odd case.

UPDATE

For $n=1$ we have $\cos(x) - \sin(x)=1$ and by squaring we get $sin(x)cos(x)=0$ which leads to $x=k\pi$ or $x=\pm \frac \pi 2 + k\pi$. From these solutions, only $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are valid.

Also $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are solutions for all odd $n$
It suffices to investigate the solutions over $[0,2\pi)$, as all solutions will be some integer multiple of $2\pi$ away from a solution in $[0,2\pi)$. Let $f(x) = \cos^n(x)-\sin^n(x)$, with $n$ odd and $n\ge 3$. Then
$$f'(x) = n(\cos^{n-1}(x)(-\sin x) - \sin^{n-1}(x)\cos(x)) = -n\sin(x)\cos(x)(\cos^{n-2}(x)+\sin^{n-2}(x)) $$
It can easily be seen that $f'$ has zeros at $0$, $\frac{\pi}{2}$, $\frac{3\pi}{4}$, $\pi$, $\frac{3\pi}{2}$, and $\frac{7\pi}{4}$, with $f'&lt;0$ on $(0,\frac{\pi}{2})\cup(\frac{3\pi}{4},\pi)\cup(\frac{3\pi}{2},\frac{7\pi}{4})$ and $f'&gt;0$ on $(\frac{\pi}{2},\frac{3\pi}{4})\cup(\pi,\frac{3\pi}{2})\cup(\frac{7\pi}{4},2\pi)$.

Now
\begin{align} f(0) &amp;= 1 \\
f(\frac{\pi}{2}) &amp;= -1 \\
f(\frac{3\pi}{4}) &amp;= -2\left(\frac{1}{\sqrt{2}}\right)^n \\
f(\pi) &amp;= -1 \\
f(\frac{3\pi}{2}) &amp;= 1 \\
f(\frac{7\pi}{4}) &amp;= 2\left(\frac{1}{\sqrt{2}}\right)^n
\end{align}
From this, we see that $0$ and $\frac{3\pi}{2}$ are solutions. Furthermore, since $f'&lt;0$ on $(0,\frac{\pi}{2})$, we have that $f(x) &lt; f(0) = 1$ for $x\in(0,\frac{\pi}{2})$. Similarly, since $f'&gt;0$ on $(\frac{\pi}{2},\frac{3\pi}{4})$, we have that $f(x)&lt;f(\frac{3\pi}{4}) = -2\left(\frac{1}{\sqrt{2}}\right)^n &lt; 1 $ for $x\in(\frac{\pi}{2},\frac{3\pi}{4})$. Applying similar reasoning to the other four intervals, we find that $f(x)&lt;1$ for all $x\in[0,2\pi)$ except $x=0$ and $x=\frac{3\pi}{2}$.

It follows that the only solutions in $[0,2\pi)$ are $x=0$ and $x=\frac{3\pi}{2}$, so the only solutions in $\mathbb{R}$ are $\boxed{x = 2k\pi}$ and $\boxed{x = \frac{3\pi}{2} + 2k\pi}$ for $k\in\mathbb{Z}$.
It will only satisfy at 2kπ where, k is an integer. Because when n is odd then at only even multiples of π theq equation will be satisfied.