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Question Number 119266 by benjo_mathlover last updated on 23/Oct/20
solve (((√3)−1)/(sin x)) + (((√3)+1)/(cos x)) = 4(√2) where x∈ (0,(π/2))
$$\:{solve}\:\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{sin}\:{x}}\:+\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${where}\:{x}\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$
Commented by MJS_new last updated on 23/Oct/20
(1+(√3))sin x −(1−(√3))cos x =4(√2)sin x cos x [sorry no time to type this transformation] 2(√2)sin (x+(π/(12))) =2(√2)sin 2x sin (x+(π/(12))) =sin 2x obviously one solution is x+(π/(12))=2x ⇒ x=(π/(12)) generally sin 2x =sin (x+(π/(12))) x=(π/(12))+2nπ∨x=((11π)/(36))+((2n)/3)π ⇒ for 0≤x<(π/2) we get x=(π/(12))∨x=((11π)/(36))
$$\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\mathrm{sin}\:{x}\:−\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)\mathrm{cos}\:{x}\:=\mathrm{4}\sqrt{\mathrm{2}}\mathrm{sin}\:{x}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\left[\mathrm{sorry}\:\mathrm{no}\:\mathrm{time}\:\mathrm{to}\:\mathrm{type}\:\mathrm{this}\:\mathrm{transformation}\right] \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{12}}\right)\:=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{12}}\right)\:=\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathrm{obviously}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{is} \\ $$$${x}+\frac{\pi}{\mathrm{12}}=\mathrm{2}{x}\:\Rightarrow\:{x}=\frac{\pi}{\mathrm{12}} \\ $$$$\mathrm{generally} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{12}}\right) \\ $$$${x}=\frac{\pi}{\mathrm{12}}+\mathrm{2}{n}\pi\vee{x}=\frac{\mathrm{11}\pi}{\mathrm{36}}+\frac{\mathrm{2}{n}}{\mathrm{3}}\pi \\ $$$$\Rightarrow\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{2}}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}=\frac{\pi}{\mathrm{12}}\vee{x}=\frac{\mathrm{11}\pi}{\mathrm{36}} \\ $$
Commented by benjo_mathlover last updated on 23/Oct/20
thank you all
$${thank}\:{you}\:{all} \\ $$
Answered by Bird last updated on 23/Oct/20
⇒((√3)−1)cosx+((√3)+1)sinx =4(√2)sinx cosx we have (√(((√3)−1)^2 +((√3)+1)^2 ))= (√(4−2(√3)+4+2(√3)))=2(√2) e⇒2(√(2{))((((√3)−1)/(2(√2))))cosx+(((√3)+1)/(2(√2)))sinx} =2(√2)sin(2x) ∃θ /sinθ=(((√3)−1)/(2(√2))) snd cosθ =(((√3)+1)/(2(√2))) ⇒θ=arctan((((√3)−1)/( (√3)+1))) ⇒sinθ cosx+cosθ sinx=sin(2x) ⇒sin(θ+x)=sin(2x) ⇒ θ+x=2x+2kπ or θ+x=π−2x+2kπ ⇒ x=θ−2kπ or 3x=π−θ+2kπ ⇒ x=θ−2kπ or x=((π−θ)/3)+((2kπ)/3)
$$\Rightarrow\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){cosx}+\left(\sqrt{\mathrm{3}}+\mathrm{1}\right){sinx} \\ $$$$=\mathrm{4}\sqrt{\mathrm{2}}{sinx}\:{cosx}\:{we}\:{have} \\ $$$$\sqrt{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left.{e}\Rightarrow\mathrm{2}\sqrt{\mathrm{2}\left\{\right.}\left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right){cosx}+\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{sinx}\right\} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\:\:\exists\theta\:/{sin}\theta=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${snd}\:{cos}\theta\:=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow\theta={arctan}\left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\right) \\ $$$$\Rightarrow{sin}\theta\:{cosx}+{cos}\theta\:{sinx}={sin}\left(\mathrm{2}{x}\right) \\ $$$$\Rightarrow{sin}\left(\theta+{x}\right)={sin}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$$\theta+{x}=\mathrm{2}{x}+\mathrm{2}{k}\pi\:{or}\:\theta+{x}=\pi−\mathrm{2}{x}+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$${x}=\theta−\mathrm{2}{k}\pi\:\:{or}\:\mathrm{3}{x}=\pi−\theta+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$${x}=\theta−\mathrm{2}{k}\pi\:{or}\:{x}=\frac{\pi−\theta}{\mathrm{3}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}} \\ $$
Answered by bemath last updated on 23/Oct/20
(((√3)−1)/(4(√2))) cos x + (((√3)+1)/(4(√2))) sin x = (1/2)sin 2x (((√3)−1)/(2(√2))) cos x + (((√3)+1)/(2(√2))) sin x = sin 2x (((√6)−(√2))/4) cos x + (((√6)+(√2))/4) sin x = sin 2x sin 15° cos x + cos 15° sin x = sin 2x sin (x+15°) = sin 2x → { ((2x = x+15° +2nπ)),((2x=180°−x−15°+2nπ)) :} → { ((x=15°)),((x=55°)) :}
$$\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\mathrm{cos}\:{x}\:+\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{cos}\:{x}\:+\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{sin}\:{x}\:=\:\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}\:\mathrm{cos}\:{x}\:+\:\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\:\mathrm{sin}\:{x}\:=\:\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathrm{sin}\:\mathrm{15}°\:\mathrm{cos}\:{x}\:+\:\mathrm{cos}\:\mathrm{15}°\:\mathrm{sin}\:{x}\:=\:\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathrm{sin}\:\left({x}+\mathrm{15}°\right)\:=\:\mathrm{sin}\:\mathrm{2}{x}\: \\ $$$$\rightarrow\begin{cases}{\mathrm{2}{x}\:=\:{x}+\mathrm{15}°\:+\mathrm{2}{n}\pi}\\{\mathrm{2}{x}=\mathrm{180}°−{x}−\mathrm{15}°+\mathrm{2}{n}\pi}\end{cases} \\ $$$$\rightarrow\begin{cases}{{x}=\mathrm{15}°}\\{{x}=\mathrm{55}°}\end{cases}\: \\ $$
Commented by MJS_new last updated on 23/Oct/20
great!
$$\mathrm{great}! \\ $$
Commented by bemath last updated on 23/Oct/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by 1549442205PVT last updated on 23/Oct/20
(((√3)−1)/(sin x)) + (((√3)+1)/(cos x)) = 4(√2).Since two sides are positive ,squaring both two sides we get the equivalent equation ((4−2(√3))/(sin^2 x))+((4+2(√3))/(cos^2 x))+(4/(sinxcosx))=32 ⇔(2−(√3))(1+(1/(tan^2 x)))+(2+(√3))(1+tan^2 )+((2(1+tan^2 x))/(tanx))=16 ⇔(((2−(√3))(1+t^2 ))/t^2 )+(2+(√3))(1+t^2 )+((2(1+t^2 ))/t)=16(with t=tanx) ⇔(2+(√3))t^4 +2t^3 −12t^2 +2t+2−(√3)=0 ⇔[t−(2−(√3))][(2+(√3))t^3 +3t^2 −3(2+(√3))t−1]=0 i)t−(2−(√3))=0⇒tanx=t=2−(√3) ⇒x=15° ii)(2+(√3))t^3 +3t^2 −3(2+(√3))t−1=0 ⇔t=1.428148 ,two negative roots rejected ⇒tanx=1.428148⇒x=55° Thus,the given equation has two roots x=15° and x=55°
$$\:\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{sin}\:{x}}\:+\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\mathrm{4}\sqrt{\mathrm{2}}.\mathrm{Since}\:\mathrm{two}\:\mathrm{sides} \\ $$$$\mathrm{are}\:\mathrm{positive}\:,\mathrm{squaring}\:\mathrm{both}\:\mathrm{two}\:\mathrm{sides} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{equivalent}\:\mathrm{equation} \\ $$$$\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{4}}{\mathrm{sinxcosx}}=\mathrm{32} \\ $$$$\Leftrightarrow\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}}\right)+\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \right)+\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{tanx}}=\mathrm{16} \\ $$$$\Leftrightarrow\frac{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{2}} }+\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)+\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}}=\mathrm{16}\left(\mathrm{with}\:\mathrm{t}=\mathrm{tanx}\right) \\ $$$$\Leftrightarrow\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{t}^{\mathrm{4}} +\mathrm{2t}^{\mathrm{3}} −\mathrm{12t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{2}−\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\Leftrightarrow\left[\mathrm{t}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right]\left[\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{t}^{\mathrm{3}} +\mathrm{3t}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{t}−\mathrm{1}\right]=\mathrm{0} \\ $$$$\left.\mathrm{i}\right)\mathrm{t}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=\mathrm{0}\Rightarrow\mathrm{tanx}=\mathrm{t}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{15}° \\ $$$$\left.\mathrm{ii}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{t}^{\mathrm{3}} +\mathrm{3t}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{t}=\mathrm{1}.\mathrm{428148}\:,\mathrm{two}\:\mathrm{negative}\:\mathrm{roots}\:\mathrm{rejected} \\ $$$$\Rightarrow\mathrm{tanx}=\mathrm{1}.\mathrm{428148}\Rightarrow\mathrm{x}=\mathrm{55}° \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{two}\:\mathrm{roots} \\ $$$$\mathrm{x}=\mathrm{15}°\:\mathrm{and}\:\mathrm{x}=\mathrm{55}° \\ $$

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