4sin²xtanx+4sin²x-3tanx-3=0

Question

It is a question of chapter no.13 exercise 13.2 question no.15

Answer 1

Let tan x = y, so sin x = y/(1+y^2):

4y^3/(1+y^2)+4y^2/(1+y^2)-3y-3=0.

Multiply through by 1+y^2:

4y^3+4y^2-3y-3y^3-3-3y^2=0;

y^3+y^2-3y-3=0;

y^3-3y+y^2-3=0=y(y^2-3)+(y^2-3)=(y+1)(y^2-3).

So y=-1 or ±√3, and x=arctan(-1) or arctan(±√3).

x=315° or 7π/4, 135º or 3π/4 (and all angles + 360n or 2πn  where n is an integer);

and x=60º or π/3, 120º or 2π/3, 240º or 4π/3, 300º or 5π/3.