f(x) = 3 + x^2 + tan(pi(x/2)), where -1 < x < 1.

Find f^-1 (3).

I got up to:

f^-1 (x) = 3 + y^2 + tan(pi(y/2)), where -1 < y < 1

then subing x= 3

I get 0 = y^2 + tan(pi(y/2))

And I'm stuck again!

What should I do with the y within the tangent

Any advice and hint is greatly appreciated ! Thanks !