Rotation.

b=0 is not the only solution here, rather can b be any integer multiple of pi.

In Mathcad's solution a could be any complex value, but IMO we have to exclude integer miltiples of pi/2 which would make one of the denominators to zero.

So the solutions for X are either k*j*2*pi (including X=0) and (2k+1)*j*pi, where k is any integer.

We can also use

or

and try to encypher Mathcad's case distinctions.

@Loi Le: Please don't put a solve block in a program structure - that can't work.