Solve-sin-1-x-sin-1-x-2-2pi-3- Tinku Tara June 3, 2023 None FacebookTweetPin Question Number 131107 by ZiYangLee last updated on 01/Feb/21 Solvesin−1x+sin−1x2=2π3 Answered by mr W last updated on 01/Feb/21 lett=x2>0sin−1(2t)=2π3−sin−1t2t=sin(2π3−sin−1t)=32cos(sin−1t)+12t3t=3cos(sin−1t)3t=3(1−t2)9t2=3(1−t2)t2=14t=12⇒x=1 Answered by EDWIN88 last updated on 02/Feb/21 ⇔sin(sin−1x+sin−1(x2))=sin2π3⇔xcos(sin−1(x2))+x2cos(sin−1(x))=123⇔x1−x24+x21−x2=123⇔x4−x2+x1−x2=3⇔x4−x2=3−x1−x2⇒x2(4−x2)=3−2x3−3x2+x2(1−x2)⇒3x2=3−2x3−3x2⇒4x2(3−3x2)=9(1−x2)2⇒leth=x2⇒4h(3−3h)=9(1−2h+h2)⇒12h−12h2=9−18h+9h2⇒4h−4h2=3−6h+3h2⇒7h2−10h+3=0⇒(7h−3)(h−1)=0⇒{h=1⇒x2=1;x=1h=37⇒x=217 Commented by mr W last updated on 02/Feb/21 x=217isnotasolutionofsin−1x+sin−1x2=2π3,butasolutionofsin−1x+sin−1x2=π3.thisisduetoyourstep1.becausesin2π3=sinπ3=12. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Evaluate-1-sin-x-dx-Next Next post: advanced-calculus-0-x-b-1-1-x-a-s-dx-b-a-s-b-a-a-s-