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Question Number 131107 by ZiYangLee last updated on 01/Feb/21

Solve sin^(−1) x+sin^(−1) (x/2)=((2π)/3)

$$\mathrm{Solve}\:\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$

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

$${let}\:{t}=\frac{{x}}{\mathrm{2}}>\mathrm{0} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \:\left(\mathrm{2}{t}\right)=\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{sin}^{−\mathrm{1}} {t} \\ $$$$\mathrm{2}{t}=\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{sin}^{−\mathrm{1}} {t}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} {t}\right)+\frac{\mathrm{1}}{\mathrm{2}}{t} \\ $$$$\mathrm{3}{t}=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} {t}\right) \\ $$$$\mathrm{3}{t}=\sqrt{\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{9}{t}^{\mathrm{2}} =\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{2}} \right) \\ $$$${t}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$

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) )) :}

$$\Leftrightarrow\:\mathrm{sin}\:\left(\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)\right)=\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\Leftrightarrow{x}\:\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)\right)+\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$\Leftrightarrow\:{x}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}\:+\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$\Leftrightarrow\:{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{3}} \\ $$$$\Leftrightarrow\:{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{3}}\:−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{4}−{x}^{\mathrm{2}} \right)=\mathrm{3}−\mathrm{2}{x}\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} \:=\:\mathrm{3}−\mathrm{2}{x}\sqrt{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} \right)=\mathrm{9}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{let}\:{h}={x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{h}\left(\mathrm{3}−\mathrm{3}{h}\right)=\mathrm{9}\left(\mathrm{1}−\mathrm{2}{h}+{h}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{12}{h}−\mathrm{12}{h}^{\mathrm{2}} =\mathrm{9}−\mathrm{18}{h}+\mathrm{9}{h}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{h}−\mathrm{4}{h}^{\mathrm{2}} =\mathrm{3}−\mathrm{6}{h}+\mathrm{3}{h}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{7}{h}^{\mathrm{2}} −\mathrm{10}{h}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{7}{h}−\mathrm{3}\right)\left({h}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{h}=\mathrm{1}\Rightarrow{x}^{\mathrm{2}} =\mathrm{1}\:;\:{x}=\mathrm{1}}\\{{h}=\frac{\mathrm{3}}{\mathrm{7}}\Rightarrow{x}=\frac{\sqrt{\mathrm{21}}}{\mathrm{7}}\:}\end{cases} \\ $$$$ \\ $$

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).

$${x}=\frac{\sqrt{\mathrm{21}}}{\mathrm{7}}\:{is}\:{not}\:{a}\:{solution}\:{of} \\ $$$$\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}=\frac{\mathrm{2}\pi}{\mathrm{3}}, \\ $$$${but}\:{a}\:{solution}\:{of} \\ $$$$\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}=\frac{\pi}{\mathrm{3}}. \\ $$$${this}\:{is}\:{due}\:{to}\:{your}\:{step}\:\mathrm{1}.\:{because} \\ $$$$\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}=\mathrm{sin}\:\frac{\pi}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$

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