Question

If $\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9}$, then find the value of $x$.

Answer 1

We have,

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

We know that

$\cos 2x=1-2{\sin }^{-1}x$

From equation (1)

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

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

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

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

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

Hence, the value of $x$ is $\pm \frac{2}{3}$