Question

If $0<{\cos }^{-1}x<1$ and $1+\sin \left({\cos }^{-1}x\right)+{\sin }^2\left({\cos }^{-1}x\right)+{\sin }^3\left({\cos }^{-1}x\right)+\dots \infty =2$, then the value of $12x^2$ is

Answer 1

$1+\sin \left({\cos }^{-1}x\right)+{\sin }^2\left({\cos }^{-1}x\right)+...\infty =2$
$\Rightarrow \frac{1}{1-\sin \left({\cos }^{-1}x\right)}=2$
or $\frac{1}{2}=1-\sin \left({\cos }^{-1}x\right)$
or $\sin \left({\cos }^{-1}x\right)=\frac{1}{2}$ or ${\cos }^{-1}x=\frac{\pi }{6}$
or $x=\frac{\sqrt{3}}{2}$ or $12x^2=9$