For,x∈12,1,sin−13x−4x3=

Let x=sinθ.  Since x∈12,1, ⇒sinθ∈12,1 ⇒π6≤θ≤π2 ⇒π2≤3θ≤3π2 ⇒-3π2≤-3θ≤-π2 ⇒π-3π2≤π-3θ≤π-π2 ⇒-π2≤π-3θ≤π2 Now sin-13x-4x3=sin-13sinθ-4sin3θ                                      =sin-1sin3θ                                      =sin-1sinπ-3θ                                       =π-3θ                                        =π-3sin-1x