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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

Thesolutionsetoftheequation4sinθcosθ−2cosθ−23sinθ+3=0intheinterval(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)

lett=tanθ2⇔θ=2arctant−8t(t2−1)(t2+1)2+2(t2−1)t2+1−43tt2+1+3=0(2+3)t4−4(2+3)t3+23t2+4(2−3)t−2+3=0t4−4t3−2(3−23)t2+4(7−43)t−7+43=0(t−2+3)2(t+2−3)(t−2−3)=0t1,2=2−3t3=−2+3t4=2+3t=tanθ2⇒x1,2=π6x3=11π6x4=5π6

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