Find the angles in a triangle

Question

In triangle $ABC$ if angle $B>C$ and both $B$ and $C$ satisfy $3\tan x-\tan^3x=k\sec^3x,(0<k<1)$ then angle A is

Answer 1

$\frac{3sinx}{cosx}- \frac{sin^3x}{cos^3x}= \frac{k}{cos^3x}$

$3sinx.cos^2x-sin^3x=k$

$sin3x=k$

i.e. $sin3B=sinC$

$3B= \pi-3C$

$B+C= \pi/3$

i.e. $A=2\pi/3$