How can I find all the solutions of $\sin^5x+\cos^3x=1$

Hint: $ \sin^5 x\leq \sin^2 x$ and $ \cos ^3 x \leq \cos^2 x $.

Hint: Pythagorean Identity for trigonometric functions.

Use what you know about the magnitudes of $\sin x$ and $\cos x$.

$$ \begin{aligned} \sin^5x+\cos^3x &\le |\sin^5x+\cos^3x| \\ &\le |\sin^5 x| + |\cos^3 x| \\ &\le |\sin^2 x| + |\cos^2 x| \\ &= \sin^2 x + \cos^2 x\\ &= 1 \end{aligned} $$

The inequality where the exponents are changed is only satisfied if the individual terms are equal: $\sin x$ and $\cos x$ must both be $0$ or $1$. Put that into the original equation, and you get $\sin x = 1$ or $\cos x = 1$. So, $x= \frac\pi2 + 2n\pi$ or $x = 2m\pi$.