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)

Commented byjohn santu last updated on 28/Jun/20

(2) x = 0

Commented byjohn 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))

Commented byjohn 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

Commented byDwaipayan 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°

Commented byDwaipayan 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))

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)}