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Solve the equation: $$ \sin^{2000}{x}+\cos^{2000}{x} =1.$$

What I did:

$\sin^2{x} =1 \land \cos^2{x}=0$ when $x=\frac \pi2 + \pi k $

$\cos^2 {x} =1 \land \sin^2{x}=0$ when $x= \pi k$

I think that these solutions apply for this equation as well but I don't really know how to formally explain it. Thanks in advance.

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    $\begingroup$ Why have you used "logical and", rather than minus, in your equations? $\endgroup$ – Glen O Apr 1 '13 at 12:05
  • $\begingroup$ I am a bit confused.Can you clarify if it is equality or inequality as you have asked to solve the inequality where as the question is an equality. $\endgroup$ – lsp Apr 1 '13 at 12:07
  • $\begingroup$ I didn't mean to use a minus. I wanted to find an x that satisfies both requirements. And yes it was a typo - it's an equation $\endgroup$ – jreing Apr 1 '13 at 12:12
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    $\begingroup$ Certainly the solutions you have found are solutions. The question is: Are there any others? Noting that $|\sin x| < 1$ and $|\cos x| < 1$ (usually) may help. $\endgroup$ – Sean Eberhard Apr 1 '13 at 12:13
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    $\begingroup$ I need a further explanation on that that last hint... You mean finding other solutions that are not a composition of 1 and 0? $\endgroup$ – jreing Apr 1 '13 at 12:19
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Hint:

You always have $$\sin^2(x)+\cos^2(x)=1$$

Now how do $\sin^2(x)+\cos^2(x)$ and $\sin^{2000}(x)+\cos^{2000}(x)$ relate if $(\sin(x),\cos(x))\neq (\pm1,0)$ and $(\sin(x),\cos(x))\neq (0,\pm1)$?

Edit (to give a complete solution after the discussion):

If $(\sin(x),\cos(x))\notin\{(\pm1,0),(0,\pm 1)\}$, then $$\sin^{2000}(x)+\cos^{2000}(x)<\sin^2(x)+\cos^2(x)=1$$ so no solution is of this form.

If $(\sin(x),\cos(x))\in\{(\pm1,0),(0,\pm 1)\}$, then the equation is clearly satisfied and you get the solutions as listed in the question.

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  • $\begingroup$ You also need to allow for (-1,0) and (0,-1). $\endgroup$ – TonyK Apr 1 '13 at 12:21
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    $\begingroup$ $sin^{2000}{x} \le sin^2{x}$ and the same for cos. (if we allow 0,1) And then... $\endgroup$ – jreing Apr 1 '13 at 12:22
  • $\begingroup$ @TonyK, thanks. Corrected it. OP: You are almost there! If the inequality is strict, then... and if it is an equality, then... $\endgroup$ – Simon Markett Apr 1 '13 at 12:29
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    $\begingroup$ so you mean that since for any non-integer $sin^{2000}{x}<sin^2{x}$ and the same for cos then $sin^{2000}{x}+cos^{2000}{x}<1$ and therefore the only solutions are the integers which I've pointed to earlier? $\endgroup$ – jreing Apr 1 '13 at 12:32
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Since we are concerned with the case when $\theta \neq k\pi/2$, we get

$$0 < \sin^2\theta < 1$$ and $$0 < \cos^2\theta < 1$$

Now observe that for $0 < \theta < \pi/2$

$$1 = (\sin^2\theta + \cos^2\theta)^2 = \sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta > \sin^4\theta + \cos^4\theta \quad (1)$$

Now observe that for $0<x<1$, $$x^n + (1-x)^n < 1$$ for $n \geq 2$, and it is monotonically strictly decreasing in n.

To show this, verify it for n=2, then use induction. The inductive step being: $$1 > x^n + (1-x)^n = (x^n + (1-x)^n)(x + 1-x) = x^{n+1} + (1-x)^{n+1} + x^n(1-x) +(1-x)^nx $$ $$ > x^{n+1} + (1-x)^{n+1}$$ Now put $x=\sin^2 \theta$ and you are done. $\theta = k\pi/2$ give the only solutions.

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