Answer
$x = -\frac{1}{2}$
Work Step by Step
$arcsin~2x+arccos~x = \frac{\pi}{6}$
$arcsin~2x= \frac{\pi}{6}-arccos~x $
$2x= sin(\frac{\pi}{6}-arccos~x)$
$2x= sin(\frac{\pi}{6})~cos(-arccos~x)+cos(\frac{\pi}{6})~sin(-arccos~x)$
$2x= (\frac{1}{2})~(x)+\frac{\sqrt{3}}{2}~(-\sqrt{1-x^2})$
$\frac{3}{2}x= \frac{\sqrt{3}}{2}~(-\sqrt{1-x^2})$
$\frac{9}{4}x^2= \frac{3}{4}~(1-x^2)$
$3x^2= \frac{3}{4}$
$x^2 = \frac{1}{4}$
$x = \pm\frac{1}{2}$
We can check our two candidate solutions:
$arcsin~2(-\frac{1}{2})+arccos~(-\frac{1}{2}) = -\frac{\pi}{2}+\frac{2\pi}{3} = \frac{\pi}{6}$
$arcsin~2(\frac{1}{2})+arccos~(\frac{1}{2}) = \frac{\pi}{2}+\frac{\pi}{3} = \frac{5\pi}{6}$
The solution is:
$x = -\frac{1}{2}$
