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

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

Answered by mr W last updated on 01/Feb/21

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

Answered by EDWIN88 last updated on 02/Feb/21

$$\iff \sin ({\sin }^{-1}x+{\sin }^{-1}\left(\frac{x}{2}\right))=\sin \frac{2\pi }{3}$$$$\iff x\cos \left({\sin }^{-1}\left(\frac{x}{2}\right)\right)+\frac{x}{2}\cos \left({\sin }^{-1}\left(x\right)\right)=\frac{1}{2}\sqrt{3}$$$$\iff x\sqrt{1-\frac{x^2}{4}}+\frac{x}{2}\sqrt{1-x^2}=\frac{1}{2}\sqrt{3}$$$$\iff x\sqrt{4-x^2}+x\sqrt{1-x^2}=\sqrt{3}$$$$\iff x\sqrt{4-x^2}=\sqrt{3}-x\sqrt{1-x^2}$$$$\Rightarrow x^2(4-x^2)=3-2x\sqrt{3-3x^2}+x^2(1-x^2)$$$$\Rightarrow 3x^2=3-2x\sqrt{3-3x^2}$$$$\Rightarrow 4x^2(3-3x^2)=9{(1-x^2)}^2$$$$\Rightarrow leth=x^2$$$$\Rightarrow 4h(3-3h)=9(1-2h+h^2)$$$$\Rightarrow 12h-12h^2=9-18h+9h^2$$$$\Rightarrow 4h-4h^2=3-6h+3h^2$$$$\Rightarrow 7h^2-10h+3=0$$$$\Rightarrow (7h-3)(h-1)=0$$$$\Rightarrow \{\begin{array}{l}h=1\Rightarrow x^2=1;x=1\\ h=\frac{3}{7}\Rightarrow x=\frac{\sqrt{21}}{7}\end{array}$$$$$$

Commented by mr W last updated on 02/Feb/21

$$x=\frac{\sqrt{21}}{7}isnotasolutionof$$$${\sin }^{-1}x+{\sin }^{-1}\frac{x}{2}=\frac{2\pi }{3},$$$$butasolutionof$$$${\sin }^{-1}x+{\sin }^{-1}\frac{x}{2}=\frac{\pi }{3}.$$$$thisisduetoyourstep1.because$$$$\sin \frac{2\pi }{3}=\sin \frac{\pi }{3}=\frac{1}{2}.$$

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