Find the conditions under which the inequality $a -\sin ( x + \arcsin (a) ) < 0$ is satisfied.

Question

For what $0<a<1$ and $0<x<\pi$, the inequality:

$a -\sin ( x + \arcsin (a) ) < 0$

is satisfied?

Edit: This is what I've tried...

$a < \sin ( x + \arcsin (a) )$ , where $|\sin ( x + \arcsin (a) )|\leq|x + \arcsin (a)|$

$a < - x - \arcsin (a)$

$ x < - (a + \arcsin(a))$

Answer 1

Your inequality is equivalent to $a < \sin (x+\sin^{-1}(a))$. This shows that if $a\geq 1$ then the inequality is never satisfied. So $0<a<1$. In such case the function $\sin^{-1}$ is injective and $\sin^{-1}(a) \in (0,\pi/2)$ which reduces the inequality to $x>0$ which is always fulfilled. Altogether I obtain $x>0$ and $0<a<1$ but I may have missed something.