Question

Solve $\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9},x>0$

Answer 1

Let ${\sin }^{-1}x=y$

$\Rightarrow$ $x=\sin y$ -------- ( 1 )

Now, $\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-x^2}$ ------- ( 2 )

Thus, the given expression will become

$\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9}$

$\Rightarrow$ $\cos 2y=\frac{1}{9}$

$\Rightarrow$ ${\cos }^{2}y-{\sin }^{2}y=\frac{1}{9}$ [ Since, $\cos 2x=co{s}^{2}x-{\sin }^{2}x$ ]

$\Rightarrow$ $1-x^2-x^2=\frac{1}{9}$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $2x^2=1-\frac{1}{9}$

$\Rightarrow$ $2x^2=\frac{8}{9}$

$\Rightarrow$ $x^2=\frac{4}{9}$

$\therefore$ $x=\pm \frac{2}{3}$

Since, $x>0$

Then, $x=\frac{2}{3}$