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Question Number 100732 by bemath last updated on 28/Jun/20

(1) If ((x+yi)/(1+i)) = (7/(7+i)) where x and y are real , what is the value of x+y (2)What are all values of x which satisfy x^2 −cos x+1 = 0 (3)What are all values of x between 0^o and 360^o which satisfy (5+2(√6))^(sin x) + (5−2(√6))^(sin x) = 2(√3)

$$\left(\mathrm{1}\right)\:\mathrm{If}\:\frac{{x}+{yi}}{\mathrm{1}+{i}}\:=\:\frac{\mathrm{7}}{\mathrm{7}+{i}}\:\mathrm{where}\:{x}\:\mathrm{and}\:\mathrm{y}\: \\ $$$$\mathrm{are}\:\mathrm{real}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}+{y}\: \\ $$$$\left(\mathrm{2}\right)\mathrm{What}\:\mathrm{are}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{which} \\ $$$$\mathrm{satisfy}\:{x}^{\mathrm{2}} −\mathrm{cos}\:{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right){W}\mathrm{hat}\:\mathrm{are}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{between}\: \\ $$$$\mathrm{0}^{\mathrm{o}} \:\mathrm{and}\:\mathrm{360}^{\mathrm{o}} \:\mathrm{which}\:\mathrm{satisfy}\: \\ $$$$\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sin}\:\mathrm{x}} \:+\:\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sin}\:\mathrm{x}} \:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Commented by john santu last updated on 28/Jun/20

(2) x = 0

$$\left(\mathrm{2}\right)\:{x}\:=\:\mathrm{0} \\ $$

Commented by john santu last updated on 28/Jun/20

(1)7x+7yi +xi −y = 7+7i (7x−y)+(x+7y)i = 7+7i { ((7x−y=7)),((x+7y=7)) :} ⇒ x =((28)/(25)) &y = ((21)/(25)) x+y = ((49)/(25))

$$\left(\mathrm{1}\right)\mathrm{7}{x}+\mathrm{7}{yi}\:+{xi}\:−{y}\:=\:\mathrm{7}+\mathrm{7}{i} \\ $$$$\left(\mathrm{7}{x}−{y}\right)+\left({x}+\mathrm{7}{y}\right){i}\:=\:\mathrm{7}+\mathrm{7}{i} \\ $$$$\begin{cases}{\mathrm{7}{x}−{y}=\mathrm{7}}\\{{x}+\mathrm{7}{y}=\mathrm{7}}\end{cases}\:\Rightarrow\:{x}\:=\frac{\mathrm{28}}{\mathrm{25}}\:\&\mathrm{y}\:=\:\frac{\mathrm{21}}{\mathrm{25}} \\ $$$${x}+{y}\:=\:\frac{\mathrm{49}}{\mathrm{25}} \\ $$

Commented by john santu last updated on 28/Jun/20

(3) 5+2(√6) = ((5+2(√6))/(5−2(√6))) × 5−2(√6) = (1/(5−2(√6))) (5+2(√6))^(sin x) + ((1/(5+2(√6))))^(sin x) = 2(√3) set (5+2(√6))^(sin x) = t ⇒t + (1/t) = 2(√3) ; t^2 −2(√3) t+1 = 0 t = ((2(√3) + (√8))/2) = (√3)+(√2) ⇒(5+2(√6))^(sin x) = (((√3)+(√2))^2 )^(1/2) ⇒(5+2(√6))^(sin x) = (5+2(√6))^(1/2) sin x = (1/2) , x = 30^o , 150^o

$$\left(\mathrm{3}\right)\:\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:=\:\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}}\:×\:\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}}\: \\ $$$$\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sin}\:\mathrm{x}} \:+\:\left(\frac{\mathrm{1}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}\right)^{\mathrm{sin}\:\mathrm{x}} =\:\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\mathrm{set}\:\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sin}\:\mathrm{x}} \:=\:\mathrm{t}\: \\ $$$$\Rightarrow\mathrm{t}\:+\:\frac{\mathrm{1}}{\mathrm{t}}\:=\:\mathrm{2}\sqrt{\mathrm{3}}\:;\:\mathrm{t}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{t}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{t}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{8}}}{\mathrm{2}}\:=\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\: \\ $$$$\Rightarrow\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sin}\:\mathrm{x}} \:=\:\left(\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sin}\:\mathrm{x}} \:=\:\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mathrm{x}\:=\:\mathrm{30}^{\mathrm{o}} ,\:\mathrm{150}^{\mathrm{o}} \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 28/Jun/20

(5+2(√6))^(sinx) +(1/((5+2(√6))^(sinx) ))=2(√3) p+(1/p)=2(√3)⇒p^2 −2(√3)p+1=0⇒ p=((2(√3)±(√(12−4)))/2)=(√3)+(√2) or(√3)−(√2) (5+2(√6))^(sinx) =(√3)+(√2) ((√3)+(√2))^(2sinx) =(√3)+(√2)⇒2sinx=1 sinx=(1/2) [x=kπ+(−1)^k (π/6)]{General solution) {k∈Z or x=30° or x=150°

$$\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{{sinx}} +\frac{\mathrm{1}}{\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{{sinx}} }=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\mathrm{p}+\frac{\mathrm{1}}{\mathrm{p}}=\mathrm{2}\sqrt{\mathrm{3}}\Rightarrow\mathrm{p}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\mathrm{p}+\mathrm{1}=\mathrm{0}\Rightarrow\:\mathrm{p}=\frac{\mathrm{2}\sqrt{\mathrm{3}}\pm\sqrt{\mathrm{12}−\mathrm{4}}}{\mathrm{2}}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\:\:\mathrm{or}\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sinx}} =\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2sinx}} =\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\Rightarrow\mathrm{2sinx}=\mathrm{1}\:\:\mathrm{sinx}=\frac{\mathrm{1}}{\mathrm{2}}\:\left[\mathrm{x}=\mathrm{k}\pi+\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\pi}{\mathrm{6}}\right]\left\{\mathrm{General}\:\mathrm{solution}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathrm{k}\in\mathbb{Z}\right. \\ $$$$\mathrm{or}\:\:\:\mathrm{x}=\mathrm{30}°\:\:\mathrm{or}\:\:\:\mathrm{x}=\mathrm{150}° \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 28/Jun/20

7x+7yi+xi−y=7+7i 7x−y+i(7y+x)=7+7i { ((7x−y=7)),((7y+x=7)) :} after solving x=((56)/(50)) y=((42)/(50))⇒ x+y=((98)/(50))

$$\:\:\: \\ $$$$\:\:\:\:\:\mathrm{7x}+\mathrm{7yi}+\mathrm{xi}−\mathrm{y}=\mathrm{7}+\mathrm{7i} \\ $$$$\:\:\:\:\mathrm{7}{x}−{y}+{i}\left(\mathrm{7}{y}+{x}\right)=\mathrm{7}+\mathrm{7}{i} \\ $$$$\:\:\:\:\:\:\begin{cases}{\mathrm{7}{x}−{y}=\mathrm{7}}\\{\mathrm{7}{y}+{x}=\mathrm{7}}\end{cases} \\ $$$${after}\:{solving}\:\:{x}=\frac{\mathrm{56}}{\mathrm{50}}\:\:{y}=\frac{\mathrm{42}}{\mathrm{50}}\Rightarrow\:{x}+{y}=\frac{\mathrm{98}}{\mathrm{50}} \\ $$

Answered by 1549442205 last updated on 28/Jun/20

1) ((x+iy)/(1+i))=(7/(7+i))⇔(7x−y−7)+(x+7y−7)i=0 ⇔ { ((7x−y−7=0(1))),((x+7y−7=0(2))) :}⇔ { ((x=((28)/(25)))),((y=((21)/(25)))) :} Hence x+y=((49)/(25)) 2)x^2 −cosx+1=0⇔x^2 +2sin^2 (x/2)=0 ⇔ { ((x^2 =0)),((sin^2 (x/2)=0)) :} ⇔x=0 3)(5+2(√6))^(sinx) +(5−2(√6))^(simx) =2(√3) ⇔((√3)+(√2))^(2sinx) +((√3)−(√2))^(2sinx) =2(√3) (1) Putting ((√3)+(√2))^(2sinx) =y⇒((√3)−(√2))^(2sinx) =(1/y)(due to ((√3)+(√2))((√3)−(√2))=1).We get the quadratic eqiation y+(1/y)=2(√3) ⇔y^2 −2(√3)y+1=0.Δ′=3−1=2,so y_1 =(√3)+(√(2 )) ,y_2 =(√3)−(√2) i)If y=(√3)+(√2) then ((√3)+(√2))^(2sinx) =(√3)+(√2) ⇔2sinx=1⇔sinx=(1/2)⇔x=(π/6)+2kπ or x=((5π)/6)+2nπ ii)If y=(√3)−(√2)=((√3)+(√2))^(−1) then ((√3)+(√2))^(2sinx) =((√3)+(√2))^(−1) ⇔2sinx=−1 ⇔sinx=((−1)/2)=sin((−π)/6)⇔x=((−π)/6)+2mπ or x=((7π)/6)+2pπ From the condition that 0<x<2π we obtain x∈{(𝛑/6);((5𝛑)/6);((7𝛑)/6)}

$$ \\ $$$$\left.\mathrm{1}\right)\:\frac{\mathrm{x}+\mathrm{iy}}{\mathrm{1}+\mathrm{i}}=\frac{\mathrm{7}}{\mathrm{7}+\mathrm{i}}\Leftrightarrow\left(\mathrm{7x}−\mathrm{y}−\mathrm{7}\right)+\left(\mathrm{x}+\mathrm{7y}−\mathrm{7}\right)\mathrm{i}=\mathrm{0} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{7x}−\mathrm{y}−\mathrm{7}=\mathrm{0}\left(\mathrm{1}\right)}\\{\mathrm{x}+\mathrm{7y}−\mathrm{7}=\mathrm{0}\left(\mathrm{2}\right)}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{28}}{\mathrm{25}}}\\{\mathrm{y}=\frac{\mathrm{21}}{\mathrm{25}}}\end{cases} \\ $$$$\mathrm{Hence}\:\mathrm{x}+\mathrm{y}=\frac{\mathrm{49}}{\mathrm{25}} \\ $$$$\left.\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{cosx}+\mathrm{1}=\mathrm{0}\Leftrightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{2sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{x}^{\mathrm{2}} =\mathrm{0}}\\{\mathrm{sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}=\mathrm{0}}\end{cases}\:\Leftrightarrow\mathrm{x}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{sinx}} +\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{simx}} =\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Leftrightarrow\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2sinx}} +\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)^{\mathrm{2sinx}} =\mathrm{2}\sqrt{\mathrm{3}}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Putting}\:\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2sinx}} =\mathrm{y}\Rightarrow\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)^{\mathrm{2sinx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{y}}\left(\mathrm{due}\:\mathrm{to}\:\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)=\mathrm{1}\right).\mathrm{We}\:\mathrm{get} \\ $$$$\mathrm{the}\:\mathrm{quadratic}\:\mathrm{eqiation}\:\mathrm{y}+\frac{\mathrm{1}}{\mathrm{y}}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{y}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\mathrm{y}+\mathrm{1}=\mathrm{0}.\Delta'=\mathrm{3}−\mathrm{1}=\mathrm{2},\mathrm{so} \\ $$$$\mathrm{y}_{\mathrm{1}} =\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}\:}\:,\mathrm{y}_{\mathrm{2}} =\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$$$\left.\mathrm{i}\right)\mathrm{If}\:\mathrm{y}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\:\mathrm{then}\:\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2sinx}} =\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{2sinx}=\mathrm{1}\Leftrightarrow\mathrm{sinx}=\frac{\mathrm{1}}{\mathrm{2}}\Leftrightarrow\mathrm{x}=\frac{\pi}{\mathrm{6}}+\mathrm{2k}\pi\:\mathrm{or} \\ $$$$\mathrm{x}=\frac{\mathrm{5}\pi}{\mathrm{6}}+\mathrm{2n}\pi \\ $$$$\left.\mathrm{ii}\right)\mathrm{If}\:\mathrm{y}=\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}=\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{−\mathrm{1}} \mathrm{then} \\ $$$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2sinx}} =\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{−\mathrm{1}} \Leftrightarrow\mathrm{2sinx}=−\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{sinx}=\frac{−\mathrm{1}}{\mathrm{2}}=\mathrm{sin}\frac{−\pi}{\mathrm{6}}\Leftrightarrow\mathrm{x}=\frac{−\pi}{\mathrm{6}}+\mathrm{2m}\pi \\ $$$$\mathrm{or}\:\mathrm{x}=\frac{\mathrm{7}\pi}{\mathrm{6}}+\mathrm{2p}\pi \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{that}\:\mathrm{0}<\mathrm{x}<\mathrm{2}\pi \\ $$$$\mathrm{we}\:\mathrm{obtain}\:\boldsymbol{\mathrm{x}}\in\left\{\frac{\boldsymbol{\pi}}{\mathrm{6}};\frac{\mathrm{5}\boldsymbol{\pi}}{\mathrm{6}};\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{6}}\right\} \\ $$$$ \\ $$$$ \\ $$

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