first thoughts $\cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x = 1 \\ \\ \\ \cos ^3 x - \sin ^3 x = \cos 3x + 3\cos x\sin ^2 x - \sin ^3 x \\ = \cos x(1 - 4\sin ^2 x + 3\sin ^2 x) - \sin ^3 x \\ = \cos (1 - \sin ^2 x) - \sin ^3 x \\ = \cos ^2 x - \sin ^3 x = 1 \\ \\ \cos ^4 x - \sin ^4 x = \cos 4x + 6\cos ^2 x\sin ^2 x - 2\sin ^4 x \\ = 1 - 8\cos ^2 x + 8\cos ^4 x + 6\cos ^2 x\sin ^2 x - 2\sin ^4 x \\ = \cos ^2 x( - 8 + 8\cos ^2 x + 6\sin ^2 x) - 2\sin ^4 x + 1 \\ = \cos ^2 x( - 8 + 2\cos ^2 x + 6) - 2\sin ^4 x + 1 \\ = \cos ^2 x(2( - 1 + \cos ^2 x)) - 2\sin ^4 x + 1 \\ = - 2\cos ^2 x\sin ^2 x - 2\sin ^4 x + 1 \\ = - 2\sin ^2 x(\sin ^2 x + \cos ^2 x) + 1 \\ = - 2\sin ^2 x + 1 = 1 \\$

The question is pretty simple, felipesa, you're making things too complicated

45 years ago IMO problems were rather different from what they are now ! If $n$ is even then $cos(x)^n = 1 + sin(x)^n$ RHS is $> 1$ except when $sin(x) = 0$ Solutions are $k \times \pi$ If $n$ is odd, same reasonning, we must have $cos(x) = 0$ and $sin(x) < 0$ or else $cos(x) > 0$ and $sin(x) = 0$ Solutions are $2 \times k \times \pi$ and $2 \times k \times \pi - \pi/2$

I solved an IMO #3 in 10 minutes Suppose $n$ is even. Then $\cos^n x$ and $\sin^n x$ are positive, so the only time $\cos^nx-\sin^nx=1$ is when $\cos x=1,\,-1$ and $\sin x=0$. This happens at $x=k\pi$ for all $k\in \mathbb{Z}$. Now suppose $n$ is odd. Then, $\cos^n x$ must be greater than or equal to $0$ (and thus $\cos x\ge0$) and $\sin^n x$ must be less than or equal to $0$ (and thus $\sin x\le 0$). This only happens in the interval $\left[\frac{3\pi}2+2k\pi,2\pi+2k\pi\right]$ for all $k\in\mathbb{Z}$. We claim that the equation is satisfied only at the endpoints. We can verify this works. If $x$ is not in the endpoints, then $\cos^nx-\sin^nx=(\cos^{n-2}x)(\cos^2x)-(\sin^{n-2}x)(1-\cos^2x)$. Since $0<\cos x<1,\,0< \cos^{n-2}x<1$, and similarly $-1< \sin^{n-2}x<0$. Thus, $(\cos^{n-2}x)(\cos^2x)-(\sin^{n-2}x)(1-\cos^2x)< (\cos^2x)-(-(1-\cos^2x))=1$, and the equation is not satisfied.