Range

2tan^-1 x = tan^-1 {2x / (1-x²)}.....if -1≤x≤1
= tan^-1 {2x / (1-x²)} + Ï€.......if x>1
= tan^-1 {2x / (1-x²)} -Ï€......if x<-1

also remember that sin2θ = 2 tan θ/(1+tan² θ)

Going by those formulae, put x = tanθ such that θ = tan^-1x.
So y = 2θ + sin^-1(2tanθ1+tan²Î¸)
=> y = 2θ + sin^-1sin2θ
=> y = 2θ + 2θ
=> y = 4θ
=> y = 4tan^-1x

Now 0 < y4 < pi2, as is the principal domain for tan^-1x.

Thus 0 < y < 2pi is the answer.

Answer (-3Ï€2,3Ï€2)

Shouldnt it be \left(-\frac{3 \pi}{4}, \frac{3 \pi}{4} \right)?

If you put x = \tan \theta, where -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}then y is given by

\begin{cases} & -(\pi + \theta) \text{ if } -\frac{\pi}{2} \le \theta \le -\frac{\pi}{4} \\ & 3 \theta \text{ if } -\frac{\pi}{4} \le \theta \le \frac{\pi}{2} \\ & \pi - \theta \text{ if } \frac{\pi}{4} \le \theta \le \frac{\pi}{2} \end{cases}