Question

Solve the following inequality:
$3-x>3\sqrt{1-x^2}.$

Answer 1

$1-x^2\ge 0$

$x^2-1\le 0$

$(x-1)(x+1)\le 0$

$xϵ[-1,1]$

$3-x>3\sqrt{1-x^2}$

Squarring both sides

$9+x^2-6x>9-9x^2$

$10x^2-6x>0$

$x(10x-6)>0$

$xϵ(-\infty ,0)\cup (\frac{3}{5},\infty )$

So, $xϵ[-1,0)\cup (\frac{3}{5},1]$