Solve for x

\hspace{-20}$Given $\bf{\sin^5 x+\cos^3 x = 1 = \sin^2 x+\cos^2 x}$\\\\\\ $\bf{\Rightarrow \sin^5 x-\sin^2 x+\cos^3 x-\cos^2 x = 1}$\\\\\\ $\bf{\Rightarrow \sin^2 x \cdot \left(\sin^3 x-1\right)+\cos^2 x\cdot (\cos x - 1) = 0}$\\\\\\ Now Using $\bf{\bullet -1\leq \sin x\;, \cos x \leq 1 }$\\\\\\ So we get $\bf{\sin^2 x\cdot \left(\sin^3 x-1\right)\leq 0}$ and $\bf{\cos^2 x\cdot (\cos x-1)\leq 0}$\\\\\\ Now equation is satisfied when $\bf{\sin^2 x\cdot \left(\sin^3 x-1\right)\;\;,\cos^2 x\cdot \left(\cos x-1\right)=0}$\\\\\\ So $\bf{\sin^2 x\cdot \left(\sin^3 x-1\right)=0\;\; \cap \;\; \cos^2 x\cdot \left(\cos x-1\right)=0}$\\\\\\ Where $\bf{\cap}$ means Intersection.....\\\\\\ $\bf{\bullet \;\; }$ If $\bf{\sin^2 x\cdot (\sin^3 x-1) = 0\;,}$ Then either $\bf{\sin^2 x= 0\Rightarrow x=n\pi}$\\\\\\ Or $\bf{\sin^3 x -1 = 0\Rightarrow \sin x -1=0\Rightarrow \sin x= 1\Rightarrow x = 2n\pi+\frac{\pi}{2}}$\\\\\\ Where $\bf{n\in \mathbb{Z}.}$\\\\\\

\hspace{-20}\bf{\bullet \;\; }$ If $\bf{\cos^2 x\cdot (\cos x-1) = 0\;,}$ Then either $\bf{\cos^2 x= 0\Rightarrow x=2n\pi\pm \frac{\pi}{2}}$\\\\\\ Or $\bf{\cos x-1 = 0\Rightarrow \cos x=1\Rightarrow x=2n\pi}$\\\\\\ Where $\bf{n\in \mathbb{Z}.}$\\\\\\ So from These $\bf{2}$ cases, we get $\bf{x = 2k\pi}$ and $\bf{x=2k\pi+\frac{\pi}{2}}$\\\\\\ Where $\bf{k\in \mathbb{Z}.}$

x can also be equal to -1/2sin^-1(2/5)

sin⁵α+cos³Î±=1
then we can say that,sin⁵α≤sin²Î± and cos³Î±â‰¤cos²Î±
here the case is in equality,then sin⁵α=sin²Î± and cos³Î±=cos²Î±
hence there are 2 cases
1> when sin²Î±=0 and cos²Î±=1
2> when sin²Î±=1 and cos²Î±=0
in 1st case the general solution is 2nπ
in 2nd case the general solution is 2kπ+π/2
[n belongs to I]