I already post a question on the solution of $$\cos^n x +\sin^n x =1,$$ but it's just a mistake. My real question is $$\cos^n x -\sin^n x =1.$$
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Please don't use the align environment in titles. – MT_ Jul 29 '16 at 00:55
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it changes nothing, see our answers – reuns Jul 29 '16 at 01:00
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Are you interested in solving for $n$ or for $x$? Is $n$ an integer? – JRN Jul 29 '16 at 01:02
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surr for x , n just is integer parameter – infox09 Jul 29 '16 at 01:05
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The problem is easy for $n$ a positive integer. For negative integers the situation is different. – André Nicolas Jul 29 '16 at 01:43
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Here's the question OP had already posted. – user236182 Oct 05 '17 at 20:10
1 Answers
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If $n$ is odd, your equation is just:
$$\cos^n (-x) + \sin^n (-x) = 1$$
reducing the problem to one already solved.
Suppose that $n$ is even. We know that $\cos^n(x) \le 1$. If $\sin (x) \neq 0$, then $\sin^n(x) > 0$, hence $\cos^n (x) - \sin^n(x) < 1$, so there are no solutions. If $\sin (x) = 0$, then $x \in \pi \Bbb Z$, so $\cos^2(x) = 1$ and the equation holds.
Summary: for even $n$, the set of solutions is $\{\pi k, k \in \Bbb Z\}$.
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yes if x is odd, the solution is clear but the other case please a small detail – infox09 Jul 29 '16 at 00:34
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