What are the sufficient conditions under which the following inequality is satisfied: $(a - (a\cos x + \sqrt{b^2-a^2}\sin x))x < 0$, where $a>0$ and $b>0$?
This is what I tried:
If $x>0$, then:
$a - (a\cos x + \sqrt{b^2-a^2}\sin x)<0$
$a < a\cos x + \sqrt{b^2-a^2}\sin x \leq |a\cos x + \sqrt{b^2-a^2}\sin x|$
$a < \sqrt{a^2 + (b^2 - a^2)}$, since $|c\cos x + d\sin x|\leq \sqrt{c^2+d^2}$
$\frac{a}{b} < 1$.
I think I've got it right until this point.
If $x<0$, then:
$- a + (a\cos x + \sqrt{b^2-a^2}\sin x)<0$
$- a < - (a\cos x + \sqrt{b^2-a^2}\sin x) \leq |a\cos x + \sqrt{b^2-a^2}\sin x|$
$- a < \sqrt{a^2 + (b^2-a^2)}$
$- a < b $
$\frac{a}{b}>-1$
Thus, to guarantee that $(a - (a\cos x + \sqrt{b^2-a^2}\sin x))x < 0$, $\forall x$, then $|\frac{a}{b}| < 1$? Is it right? Is there a more elegant and/or succint derivation?
Edit: $(a, b, x)$ are all real numbers.