$ \sin^{2000}{x}+\cos^{2000}{x} =1$ equation explanation 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. 
 A: 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.
A: 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.
