How to solve $\sin^{-1}(x)=\sin^{-1}(2/5) +\sin^{-1}(3/5)$?

Question

I tried to solve this equation:

$\sin^{-1}(x)=\sin^{-1}(2/5) +\sin^{-1}(3/5)$.

I applied the $\sin$ function on both sides and I've got

$x=\sin\left(\sin^{-1}( \frac{2}{5}) +\sin^{-1}( \frac{3}{5})\right)$.

Answer 1

Note

$$x=\sin(\sin^{-1}x)=\sin\left(\sin^{-1}\frac25+\sin^{-1}\frac35\right)$$ $$=\sin(\sin^{-1}\frac25)\cos(\sin^{-1}\frac35) +\cos(\sin^{-1}\frac25)\sin(\sin^{-1}\frac35)$$ $$=\frac25\cdot\frac45+\frac{\sqrt{21}}{5}\cdot\frac35$$