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For $0 < x < 2\pi$ and positive even $n$, the only solution for $\cos^n x-\sin^n x=1$ is $\pi$.
The argument is simple as $0\le\cos^n x, \sin^n x\le1$ and hence $\cos^n x-\sin^n x=1$ iff $\cos^n x=1$ and $\sin^n x=0$.

My question is that any nice argument to show the following statement?

'For $0 < x < 2\pi$ and positive odd $n$, the only solution for $\cos^n x-\sin^n x=1$ is $\frac{3\pi}{2}$.'

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    http://math.stackexchange.com/questions/179778/solve-cosnx-sinnx-1-with-n-in-mathbbn2012-10-24
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    @lab bhattacharjee Thanks for informing. Nevertheless, I think the solution provided here is **elegant**.2012-10-24
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    ya, it's exquisite.2012-10-24
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    This problem is IMO $1961$ Problem $3$. [Here's an AoPS thread about it.](https://artofproblemsolving.com/community/c6h55221p343304)2017-10-05
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    Here are some relevant Math.SE question links: [(1)](https://math.stackexchange.com/questions/1940444/solve-the-equation-cosnx-sinnx-1?noredirect=1&lq=1), [(2)](https://math.stackexchange.com/questions/1874442/solve-cosn-x-sinn-x-1?noredirect=1&lq=1), [(3)](https://math.stackexchange.com/questions/1874477/solve-cosn-x-sinnx-1), [(4)](https://math.stackexchange.com/questions/348085/sin2000x-cos2000x-1-equation-explanation?noredirect=1&lq=1).2017-10-05
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    https://socratic.org/questions/what-is-the-solution-for-cos-nx-sin-nx-1-witn-n-in-nn2017-10-05

1 Answers 1

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We leave the case $n = 1$ and $n = 2$ separately, and assume $n \geq 3$ from now on.

Observe that if $|r| \leq 1$, then $|r^n| \leq r^2$ with equality if and only if $r = 0$ or $|r| = 1$. Then it follows that

$$1 = \left|\cos^n x - \sin^n x\right| \leq \left|\cos^n x\right| + \left|\sin^n x\right| \leq \cos^2 x + \sin^2 x = 1. $$

This forces every intermediate inequality to be equality. In particular, we must have

$$ \cos x , \sin x \in \{0, \pm 1\}.$$

Thus $x \in \{ \frac{\pi}{2}, \pi, \frac{3\pi}{2} \}$. Now the rest is clear.

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    This is a nice solution, using the same technique to solve all the cases. Thanks!2012-10-24
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    Using this technique, $\cos^n x \pm \sin^n x=1$ can be easily solve for all positive integers $n$.2012-10-24