Solve for x: $\sin^{-1}(\sin(\frac{2x^2+4}{x^2+1}))<\pi -3$
My approach : $\pi -3$ is approximately equal to $\frac{\pi}{22.42}$ and $\frac{2x^2+4}{x^2+1}$ is positive as no condition is imposed on x we can take any value but i am not able to proceed
Hint: You know $\sin^{-1}(\sin u)=u$ then solve $$\sin^{-1}(\sin(\frac{2x^2+4}{x^2+1}))=\frac{2x^2+4}{x^2+1}=2+\frac{2}{x^2+1}<\pi -3$$