Answer
$x = \frac{\sqrt{5}}{5}$
Work Step by Step
$arcsin~2x+arcsin~x = \frac{\pi}{2}$
We can use the following identity:
$arcsin~a+arcsin~b = arcsin(a\sqrt{1-b^2}+b\sqrt{1-a^2})$
We can find the value of $x$:
$arcsin~2x+arcsin~x = \frac{\pi}{2}$
$arcsin~(2x\sqrt{1-x^2}+x\sqrt{1-(2x)^2}) = \frac{\pi}{2}$
$arcsin~(2x\sqrt{1-x^2}+x\sqrt{1-4x^2}) = \frac{\pi}{2}$
$2x\sqrt{1-x^2}+x\sqrt{1-4x^2} = 1$
$2x\sqrt{1-x^2} = 1 - x\sqrt{1-4x^2}$
$4x^2~(1-x^2) = 1 - 2x\sqrt{1-4x^2}+x^2(1-4x^2)$
$4x^2-4x^4 = 1 - 2x\sqrt{1-4x^2}+x^2-4x^4$
$3x^2 = 1 - 2x\sqrt{1-4x^2}$
$3x^2 - 1 = - 2x\sqrt{1-4x^2}$
$9x^4 - 6x^2+1 = 4x^2(1-4x^2)$
$9x^4 - 6x^2+1 = 4x^2-16x^4$
$25x^4 - 10x^2+1 = 0$
$(5x^2-1)^2 = 0$
$5x^2-1 = 0$
$x^2 = \frac{1}{5}$
$x = \pm \frac{1}{\sqrt{5}}$
$x = \pm \frac{\sqrt{5}}{5}$
We can test both of these candidate solutions in the original equation:
$arcsin~[2(- \frac{\sqrt{5}}{5})]+arcsin~(- \frac{\sqrt{5}}{5}) = -\frac{\pi}{2}$
$arcsin~[2(\frac{\sqrt{5}}{5})]+arcsin~(\frac{\sqrt{5}}{5}) = \frac{\pi}{2}$
Therefore:
$x = \frac{\sqrt{5}}{5}$
