(arcsin(x/2)+arccos(x/2))/arctan(x) = 3/2, solve for x if x >0. Thanks.

Answer 1

By definition, arcsin(x/2) would be the angle 𝜃, so that sin𝜃 = x/2

Using a hunch that perhaps we would deal with one of our standard angles,
I tried x = 0, 1, √2, √3, 2 , (which would be sides in our standard 45-45-90 and 30-60-90 triangles
for x = 0 ,
(arcsin(0)+arccos(0))/arctan(0) = (0 + 1)/0 ≠ 3/2
for x= 1
(arcsin(1/2)+arccos(1/2)/arctan(2) = (30 + 60)/63.43.. ≠ 3/2
x = √2
(arcsin(√2/2)+arccos(√2/2)/arctan(√2) = (45+45)/54.7... ≠ 3/2
x = √3
(arcsin(√3/2)+arccos(√3/2)/arctan(√3) = (60 + 30)/60 = 90/60 = 3/2
Well, how is that for a "lucky" guess

At the moment, I can't think of a formal way to solve this.