In a triangle ABC, angle A is greater than B, if the measures of angle A and B satisfy the equation3sinx−4sin3x−k=0, 0

Given A>B and, A and B satisfy the equation           3sinx−4sin3x−k=0⇒sin3x-k=0           (or)  sin3A−k=0  and  sin3B−K=0                         ∴sin3A=sin3B=sinπ-3B ⇒                          3A=π−3B    ∵A>B⇒                           A+B=π3 Then C=π−π3=2π3