Complex numbers

hint : just rationalise lhs
rhs : only possible values =±1
ans : n=2

applying am gm for x² and 1 (since given that x≥0) => x²+1 ≥ 2x but since sin^-1 is there , the only possibility is equality .:. rhs = 2pi sin^-1(1) = 1

and on rationalisation lhs = iⁿ .:. minimum n = 4

Plz check if this solution is rite??

[(i+1)²i²-1²]ⁿ=2Ï€sin^-1(1+x²2x)
==> iⁿ=2Ï€sin^-1(1+x²2x)

let x=tanθ
==>sin^-1(1+x²2x)=sin^-1(sec²Î¸2tanθ)
=sin^-11sin2θ
=sin^-1cosec2θ

since the only possible values of cosec θ for which it can satisfy sin^-1θ are ±1 and x≥0 hence the only possible value of sin θ is Ï€2 hence iⁿ=1
==>n=4