MHB Solve Trigonometry Challenge: $\cos^k x-\sin^k x=1$

[anemone]

Solve the equation $\cos^k x-\sin^k x=1$, where $k$ is a given positive integer.

 

[kaliprasad]

Solve the equation $\cos^k x-\sin^k x=1$, where $k$ is a given positive integer.

partial solution easy part(there is a difficult part also)

Spoiler

for k even.

only solution

$\cos^k x=1$
$\sin^k x=0$

so
$\sin\, x=0$ and $\cos\, x=\pm 1$ hence $x = n\pi$

k is odd

there are 3 cases

case 1)
$\cos\ x=1$
$\sin\ x=0$

this case is a subset of above case $x = 2n\pi$

case 2)
$\cos\ x=0$
$\sin\ x=-1$

giving $x = (2n-\dfrac{1}{2})\pi$

case 3)
$\cos\ x$, $\sin\ x$ are non integers

This is the difficult part(remaining )

 

[Opalg]

Solve the equation $\cos^k x-\sin^k x=1$, where $k$ is a given positive integer.

[sp]
Let $f_k(x) = \cos^k x-\sin^k x$. This is periodic, with period $2\pi$, so it will be sufficient to find solutions of $f_k(x) = 1$ in the interval $x\in [0,2\pi).$

If $k=1$ then $f_1(x) = \cos x - \sin x = \sqrt2\cos\bigl(x + \frac\pi4\bigr)$. That takes the value $1$ when $\cos\bigl(x + \frac\pi4\bigr) = \frac1{\sqrt2}$, in other words $x + \frac\pi4 = \frac\pi4 \text{ or }\frac{7\pi}4.$ Thus the solutions to $f_1(x) = 1$ are $x=0,\frac{3\pi}2.$

The case $k=1$ is anomalous, because that is the only value of $k$ for which $|\,f_k(x)|$ can ever be greater than $1$. One way to see this is by calculus, using the method of Lagrange multipliers. Suppose that $k\geqslant2$. If we let $u = \cos x$ and $v = \sin x$, then we want to find the extreme values of $u^k - v^k$ subject to the constraint $u^2+v^2 = 1$. The Lagrange method gives the equations $$ku^{k-1} - 2\lambda u = 0, \qquad -kv^{k-1} - 2\lambda v = 0.$$ Therefore either $u=0$ or $u^{k-2} = \frac{2\lambda}k$, and similarly either $v=0$ or $v^{k-2} = -\frac{2\lambda}k$. Combining those values with the fact that $u^2+v^2=1$, it follows that if $u=0$ then $v = \pm1$, and if $v=0$ then $u = \pm1$. If both $u$ and $v$ are nonzero then $u^{k-2} = -v^{k-2}$. In the case when $k$ is odd, the $(k-2)$-th roots are unique and it follows that $u=-v$, so that $(u,v) = \pm\bigl(\frac1{\sqrt2},-\frac1{\sqrt2}\bigr)$. But when $k$ is even there are two (real) $(k-2)$-th roots, so there is the additional possibility that $u=v=\pm\frac1{\sqrt2}$.

This shows that the only possible extreme values of $(u,v)$ are $(\pm1,0)$, $(0,\pm1)$ and $\bigl(\pm\frac1{\sqrt2}, \pm\frac1{\sqrt2}\bigr)$. In terms of $x$, with $(u,v) = (\cos x,\sin x)$, this means that the only extreme values of $f_k(x)$ must occur when $x$ is a multiple of $\frac\pi4$. It is easy to check that $|\,f_k(x)| \leqslant1$ at all those points (and hence everywhere), and that the only points in $[0,2\pi)$ where $f_k(x) =1$ are $x=0$, together with $x=\pi$ (if $k$ is even) and $x = \frac{3\pi}2$ (if $k$ is odd). So the solutions found by kaliprasad are the only ones.
[/sp]

 

[anemone]

Thank you Kali (for your partial solution) and Opalg for your solution that complements Kali's!

I can't wait any longer to share the other solution that I found somewhere online and I certainly hope you like the method used to solve it as much as I do:

Spoiler

For $k\ge 2$, we have

$1=\cos^k x-\sin^k x \le |cos^k x-\sin^k x|\le |\cos^k x|+|\sin^k x| \le \cos ^2 x+\sin^2 x=1$

Hence, $\sin^2 x=|\sin^k x|$ and $\cos^2 x=|\cos^k x|$, from which it follows that $\sin x,\,\cos x\in\{1,\,0,\,-1\}\implies x\in\dfrac{\pi Z}{2}$. By inspection one obtains the set of solutions

$\{n\pi,\, n\in Z\}$ for even $k$ and $\{2n\pi,\,2n\pi-\dfrac{\pi}{2},\,n\in Z\}$ for odd $k$.

For $k=1$, we have $1=\cos x-\sin x=-\sqrt{2}\sin\left(x-\dfrac{\pi}{4}\right)$ which yields the set of solutions $\{2n\pi,\,2n\pi-\dfrac{\pi}{2},\,n\in Z\}$.

Therefore, the solutions to the given equation is $\{n\pi,\,2n\pi-\dfrac{\pi}{2},\,n\in Z\}$.