2022/028) Solve $\sin ^5 x + \cos^5 x = 1$

 We know as $-1 \le \sin\, x \le 1$ so $sin^5 x \ le sin ^2 x$. they are same when sin x = 0 or 1 otherwise  $\sin^5 x \lt sin ^2 x$

similarly $\cos^5 x \le \cos ^2 x$. they are same when cos  x = 0 or 1 otherwise  $cos^5 x \ lt cos  ^2 x$

so $\sin ^5 x + \cos^5 x \le \sin ^2 x + \cos^2 x$ or 1 they are same when $\sin \, x =$ 0 or and $\cos\,x$ = 0 or 1 that is one of them is zero and another 1

this is possible when $\sin\, x = 0$ and $\cos\,x = 1$ that is $x = 2n\pi$

or $\sin\, x = 1$ and $\cos\,x = 0$ that is $x = (2n + \frac{1}{2}) \pi$

hence $x \in \{ 2n\pi, (2n + \frac{1}{2}) \pi \}$