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

Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.

$\sin^5x\le \sin^2x, \cos^3x\le \cos^2x$, and $\sin^2x+\cos^2x=1$, so $\sin^5x+\cos^3x$
can be equal to 1 if and only if $$\sin^5x=\sin^2x$$ and $$\cos^3x=\cos^2x$$, i.e.
$$\sin^2x(1-\sin^3x)=0$$ and $$\cos^2x(1-\cos x)=0$$
The first equation has solution: $x=k\pi$ or $x=\pi/2+2m\pi, k,m $ integers.
The second equation has solution: $x=2n\pi$ or $x=\pi/2+p\pi, n,p$ integers.
Therefore the solution is either:
(i) $x=k\pi$ and $x=2n\pi$, or equivalently, $x=2n\pi$ for some integer $n$, or 
(ii) $x=k\pi$ and $x=\pi/2+p\pi$ (impossible), or
(iii) $x=\pi/2+2m\pi$ and $x=2n\pi$ (impossible) or
(iv) $x=\pi/2+2m\pi$ and $x=\pi/2+p\pi$, or equivalently, $x=\pi/2+2m\pi$ for some integer $p$.
In conclusion, $x=2n\pi$ or $x=\pi/2+2n\pi$, $n$ integers, are the solutions.
