I Solved it in a different way.
First of all it's obvious that point P and Q creates two triangles with the x and y Axis.
I drew thus triangles and name the new angles respectively a & b.
we know that segment OA is √3 and PA is 1, consequently PO must be 2.
Since PO and QO are both radius of the half circle drew in the picture, they must both be 2.
since that is the classic 1: √3 : 2 right triangle that means that angle pOa must be 30 degree.
Since pOa + pOq + qOb = 180 -> 30 + 90 + qOb =180 -> qOb = 60°
Since It's obvious that qBo = 90
We now know that also riangle QBO is a right 30:60:90 Triangle,and since
qBo = 90° and QO 2
qOb = 60°, THEN Segment QB = √3
oQb = 30, then OB = 1
So the x coordinate of the point Q is s=
1
