Question Number 113986 by deepraj123 last updated on 16/Sep/20

The solution set of the equation  4 sin θ cos θ−2 cos θ−2 (√3) sin θ+(√3) =0  in the interval  (0, 2π)  is

Answered by MJS_new last updated on 16/Sep/20

let t=tan (θ/2) ⇔ θ=2arctan t  −((8t(t^2 −1))/((t^2 +1)^2 ))+((2(t^2 −1))/(t^2 +1))−((4(√3)t)/(t^2 +1))+(√3)=0  (2+(√3))t^4 −4(2+(√3))t^3 +2(√3)t^2 +4(2−(√3))t−2+(√3)=0  t^4 −4t^3 −2(3−2(√3))t^2 +4(7−4(√3))t−7+4(√3)=0  (t−2+(√3))^2 (t+2−(√3))(t−2−(√3))=0  t_(1, 2) =2−(√3)  t_3 =−2+(√3)  t_4 =2+(√3)  t=tan (θ/2)  ⇒ x_(1, 2) =(π/6)     x_3 =((11π)/6)     x_4 =((5π)/6)