(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)


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

$(1) If \frac{x + y i}{1 + i} = \frac{7}{7 + i} where x and y$ $are real, what is the value of x + y$ $(2) What are all values of x which$ $satisfy x^{2} - \cos x + 1 = 0$ $(3) W hat are all values of x between$ $0^{o} and 360^{o} which satisfy$ ${(5 + 2 \sqrt{6})}^{\sin x} + {(5 - 2 \sqrt{6})}^{\sin x} = 2 \sqrt{3}$
Commented by john santu last updated on 28/Jun/20

$(2) x = 0$
Commented by john santu last updated on 28/Jun/20

$(1) 7 x + 7 y i + x i - y = 7 + 7 i$ $(7 x - y) + (x + 7 y) i = 7 + 7 i$ ${\begin{cases} 7 x - y = 7 \\ x + 7 y = 7 \end{cases} \Rightarrow x = \frac{28}{25} & y = \frac{21}{25}$ $x + y = \frac{49}{25}$
Commented by john santu last updated on 28/Jun/20

$(3) 5 + 2 \sqrt{6} = \frac{5 + 2 \sqrt{6}}{5 - 2 \sqrt{6}} \times 5 - 2 \sqrt{6}$ $= \frac{1}{5 - 2 \sqrt{6}}$ ${(5 + 2 \sqrt{6})}^{\sin x} + {(\frac{1}{5 + 2 \sqrt{6}})}^{\sin x} = 2 \sqrt{3}$ $set {(5 + 2 \sqrt{6})}^{\sin x} = t$ $\Rightarrow t + \frac{1}{t} = 2 \sqrt{3}; t^{2} - 2 \sqrt{3} t + 1 = 0$ $t = \frac{2 \sqrt{3} + \sqrt{8}}{2} = \sqrt{3} + \sqrt{2}$ $\Rightarrow {(5 + 2 \sqrt{6})}^{\sin x} = {({(\sqrt{3} + \sqrt{2})}^{2})}^{\frac{1}{2}}$ $\Rightarrow {(5 + 2 \sqrt{6})}^{\sin x} = {(5 + 2 \sqrt{6})}^{\frac{1}{2}}$ $\sin x = \frac{1}{2}, x = 30^{o}, 150^{o}$
Commented by Dwaipayan Shikari last updated on 28/Jun/20
$(5+2(√6))^(sinx) +(1/((5+2(√6))^(sinx) ))=2(√3) p+(1/p)=2(√3)⇒p^2 −2(√3)p+1=0⇒ p=((2(√3)±(√(12−4)))/2)=(√3)+(√2) or(√3)−(√2) (5+2(√6))^(sinx) =(√3)+(√2) ((√3)+(√2))^(2sinx) =(√3)+(√2)⇒2sinx=1 sinx=(1/2) [x=kπ+(−1)^k (π/6)]{General solution) {k∈Z or x=30° or x=150°$
${(5 + 2 \sqrt{6})}^{s i n x} + \frac{1}{{(5 + 2 \sqrt{6})}^{s i n x}} = 2 \sqrt{3}$ $p + \frac{1}{p} = 2 \sqrt{3} \Rightarrow p^{2} - 2 \sqrt{3} p + 1 = 0 \Rightarrow p = \frac{2 \sqrt{3} \pm \sqrt{12 - 4}}{2} = \sqrt{3} + \sqrt{2} or \sqrt{3} - \sqrt{2}$ ${(5 + 2 \sqrt{6})}^{sinx} = \sqrt{3} + \sqrt{2}$ ${(\sqrt{3} + \sqrt{2})}^{2 sinx} = \sqrt{3} + \sqrt{2} \Rightarrow 2 sinx = 1 sinx = \frac{1}{2} [x = k π + {(- 1)}^{k} \frac{π}{6}] {General solution)$ ${k \in Z$ $or x = 30 ° or x = 150 °$
Commented by Dwaipayan Shikari last updated on 28/Jun/20

$7 x + 7 yi + xi - y = 7 + 7 i$ $7 x - y + i (7 y + x) = 7 + 7 i$ ${\begin{cases} 7 x - y = 7 \\ 7 y + x = 7 \end{cases}$ $a f t e r s o l v i n g x = \frac{56}{50} y = \frac{42}{50} \Rightarrow x + y = \frac{98}{50}$
	Answered by 1549442205 last updated on 28/Jun/20
	$1) ((x+iy)/(1+i))=(7/(7+i))⇔(7x−y−7)+(x+7y−7)i=0 ⇔ { ((7x−y−7=0(1))),((x+7y−7=0(2))) :}⇔ { ((x=((28)/(25)))),((y=((21)/(25)))) :} Hence x+y=((49)/(25)) 2)x^2 −cosx+1=0⇔x^2 +2sin^2 (x/2)=0 ⇔ { ((x^2 =0)),((sin^2 (x/2)=0)) :} ⇔x=0 3)(5+2(√6))^(sinx) +(5−2(√6))^(simx) =2(√3) ⇔((√3)+(√2))^(2sinx) +((√3)−(√2))^(2sinx) =2(√3) (1) Putting ((√3)+(√2))^(2sinx) =y⇒((√3)−(√2))^(2sinx) =(1/y)(due to ((√3)+(√2))((√3)−(√2))=1).We get the quadratic eqiation y+(1/y)=2(√3) ⇔y^2 −2(√3)y+1=0.Δ′=3−1=2,so y_1 =(√3)+(√(2 )) ,y_2 =(√3)−(√2) i)If y=(√3)+(√2) then ((√3)+(√2))^(2sinx) =(√3)+(√2) ⇔2sinx=1⇔sinx=(1/2)⇔x=(π/6)+2kπ or x=((5π)/6)+2nπ ii)If y=(√3)−(√2)=((√3)+(√2))^(−1) then ((√3)+(√2))^(2sinx) =((√3)+(√2))^(−1) ⇔2sinx=−1 ⇔sinx=((−1)/2)=sin((−π)/6)⇔x=((−π)/6)+2mπ or x=((7π)/6)+2pπ From the condition that 0<x<2π we obtain x∈{(𝛑/6);((5𝛑)/6);((7𝛑)/6)}$
	$1) \frac{x + iy}{1 + i} = \frac{7}{7 + i} \Leftrightarrow (7 x - y - 7) + (x + 7 y - 7) i = 0$ $\Leftrightarrow {\begin{cases} 7 x - y - 7 = 0 (1) \\ x + 7 y - 7 = 0 (2) \end{cases} \Leftrightarrow {\begin{cases} x = \frac{28}{25} \\ y = \frac{21}{25} \end{cases}$ $Hence x + y = \frac{49}{25}$ $2) x^{2} - cosx + 1 = 0 \Leftrightarrow x^{2} + {2 \sin}^{2} \frac{x}{2} = 0$ $\Leftrightarrow {\begin{cases} x^{2} = 0 \\ \sin^{2} \frac{x}{2} = 0 \end{cases} \Leftrightarrow x = 0$ $3) {(5 + 2 \sqrt{6})}^{sinx} + {(5 - 2 \sqrt{6})}^{simx} = 2 \sqrt{3}$ $\Leftrightarrow {(\sqrt{3} + \sqrt{2})}^{2 sinx} + {(\sqrt{3} - \sqrt{2})}^{2 sinx} = 2 \sqrt{3} (1)$ $Putting {(\sqrt{3} + \sqrt{2})}^{2 sinx} = y \Rightarrow {(\sqrt{3} - \sqrt{2})}^{2 sinx}$ $= \frac{1}{y} (due to (\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2}) = 1) . We get$ $the quadratic eqiation y + \frac{1}{y} = 2 \sqrt{3}$ $\Leftrightarrow y^{2} - 2 \sqrt{3} y + 1 = 0 . Δ^{'} = 3 - 1 = 2, so$ $y_{1} = \sqrt{3} + \sqrt{2}, y_{2} = \sqrt{3} - \sqrt{2}$ $i) If y = \sqrt{3} + \sqrt{2} then {(\sqrt{3} + \sqrt{2})}^{2 sinx} = \sqrt{3} + \sqrt{2}$ $\Leftrightarrow 2 sinx = 1 \Leftrightarrow sinx = \frac{1}{2} \Leftrightarrow x = \frac{π}{6} + 2 k π or$ $x = \frac{5 π}{6} + 2 n π$ $ii) If y = \sqrt{3} - \sqrt{2} = {(\sqrt{3} + \sqrt{2})}^{- 1} then$ ${(\sqrt{3} + \sqrt{2})}^{2 sinx} = {(\sqrt{3} + \sqrt{2})}^{- 1} \Leftrightarrow 2 sinx = - 1$ $\Leftrightarrow sinx = \frac{- 1}{2} = \sin \frac{- π}{6} \Leftrightarrow x = \frac{- π}{6} + 2 m π$ $or x = \frac{7 π}{6} + 2 p π$ $From the condition that 0 < x < 2 π$ $we obtain x \in {\frac{π}{6}; \frac{5 π}{6}; \frac{7 π}{6}}$
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