Solve for x: $\sin^{-1}(\sin (\frac{2x^2+4}{x^2+1}))<\pi -3$

Question

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

score 2 · Answer 1 · answered Oct 14 '18 at 13:00

2

Hint: You know $\sin^{-1}(\sin u)=u$ then solve $$\sin^{-1}(\sin(\frac{2x^2+4}{x^2+1}))=\frac{2x^2+4}{x^2+1}=2+\frac{2}{x^2+1}<\pi -3$$

answered Oct 14 '18 at 13:00

Nosrati

29,995

It's only true that $\sin^{-1}(\sin(u)) = u$ if $u$ is between $-\pi/2$ and $\pi/2$, though. – hunter Oct 14 '18 at 13:02
yes, of course. $\pi-3<\pi/2$. – Nosrati Oct 14 '18 at 13:03

Solve for x: $\sin^{-1}(\sin (\frac{2x^2+4}{x^2+1}))<\pi -3$

1 Answers1