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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
$$\mathrm{The}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{4}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta−\mathrm{2}\:\mathrm{cos}\:\theta−\mathrm{2}\:\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\:=\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\:\left(\mathrm{0},\:\mathrm{2}\pi\right)\:\:\mathrm{is} \\ $$
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)
$$\mathrm{let}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:\Leftrightarrow\:\theta=\mathrm{2arctan}\:{t} \\ $$$$−\frac{\mathrm{8}{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{2}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{4}\sqrt{\mathrm{3}}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{4}} −\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{3}} +\mathrm{2}\sqrt{\mathrm{3}}{t}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){t}−\mathrm{2}+\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$${t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} −\mathrm{2}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right){t}−\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left({t}+\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({t}−\mathrm{2}−\sqrt{\mathrm{3}}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1},\:\mathrm{2}} =\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${t}_{\mathrm{3}} =−\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${t}_{\mathrm{4}} =\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}_{\mathrm{1},\:\mathrm{2}} =\frac{\pi}{\mathrm{6}}\:\:\:\:\:{x}_{\mathrm{3}} =\frac{\mathrm{11}\pi}{\mathrm{6}}\:\:\:\:\:{x}_{\mathrm{4}} =\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$

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