Question

Find the value of $x$ if ${\sin }^{-1}\left(\frac{2}{3}\right)+{\sin }^{-1}\left(\frac{2}{3}\right)={\sin }^{-1}x$

Answer 1

We have,

${\sin }^{-1}\left(\frac{2}{3}\right)+{\sin }^{-1}\left(\frac{2}{3}\right)={\sin }^{-1}x$

$\Rightarrow 2{\sin }^{-1}\left(\frac{2}{3}\right)={\sin }^{-1}x\therefore 2{\sin }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

$\Rightarrow {\sin }^{-1}(2\times \frac{2}{3}\sqrt{1-{\left(\frac{2}{3}\right)}^2})={\sin }^{-1}x$

$\Rightarrow \frac{4}{3}\sqrt{1-\frac{4}{9}}=x$

$\Rightarrow x=\frac{4}{3}\sqrt{\frac{5}{3^2}}$

$\Rightarrow x=\frac{4}{9}\sqrt{5}$

Hence, this is the answer.