Menu Close

Solve-sin-1-x-sin-1-x-2-2pi-3-




Question Number 131107 by ZiYangLee last updated on 01/Feb/21
Solve sin^(−1) x+sin^(−1) (x/2)=((2π)/3)
Solvesin1x+sin1x2=2π3
Answered by mr W last updated on 01/Feb/21
let t=(x/2)>0  sin^(−1)  (2t)=((2π)/3)−sin^(−1) t  2t=sin (((2π)/3)−sin^(−1) t)=((√3)/2) cos (sin^(−1) t)+(1/2)t  3t=(√3) cos (sin^(−1) t)  3t=(√(3(1−t^2 )))  9t^2 =3(1−t^2 )  t^2 =(1/4)  t=(1/2)  ⇒x=1
lett=x2>0sin1(2t)=2π3sin1t2t=sin(2π3sin1t)=32cos(sin1t)+12t3t=3cos(sin1t)3t=3(1t2)9t2=3(1t2)t2=14t=12x=1
Answered by EDWIN88 last updated on 02/Feb/21
⇔ sin (sin^(−1) x+sin^(−1) ((x/2)))=sin ((2π)/3)  ⇔x cos (sin^(−1) ((x/2)))+(x/2)cos (sin^(−1) (x))= (1/2)(√3)  ⇔ x(√(1−(x^2 /4))) +(x/2)(√(1−x^2 )) = (1/2)(√3)  ⇔ x(√(4−x^2 )) +x(√(1−x^2 )) = (√3)  ⇔ x(√(4−x^2 )) = (√3) −x(√(1−x^2 ))  ⇒x^2 (4−x^2 )=3−2x(√(3−3x^2 ))+x^2 (1−x^2 )  ⇒3x^2  = 3−2x(√(3−3x^2 ))  ⇒4x^2 (3−3x^2 )=9(1−x^2 )^2   ⇒let h=x^2   ⇒4h(3−3h)=9(1−2h+h^2 )  ⇒12h−12h^2 =9−18h+9h^2   ⇒4h−4h^2 =3−6h+3h^2   ⇒7h^2 −10h+3=0  ⇒(7h−3)(h−1)=0  ⇒ { ((h=1⇒x^2 =1 ; x=1)),((h=(3/7)⇒x=((√(21))/7) )) :}
sin(sin1x+sin1(x2))=sin2π3xcos(sin1(x2))+x2cos(sin1(x))=123x1x24+x21x2=123x4x2+x1x2=3x4x2=3x1x2x2(4x2)=32x33x2+x2(1x2)3x2=32x33x24x2(33x2)=9(1x2)2leth=x24h(33h)=9(12h+h2)12h12h2=918h+9h24h4h2=36h+3h27h210h+3=0(7h3)(h1)=0{h=1x2=1;x=1h=37x=217
Commented by mr W last updated on 02/Feb/21
x=((√(21))/7) is not a solution of  sin^(−1) x+sin^(−1) (x/2)=((2π)/3),  but a solution of  sin^(−1) x+sin^(−1) (x/2)=(π/3).  this is due to your step 1. because  sin ((2π)/3)=sin (π/3)=(1/2).
x=217isnotasolutionofsin1x+sin1x2=2π3,butasolutionofsin1x+sin1x2=π3.thisisduetoyourstep1.becausesin2π3=sinπ3=12.