How do I solve the inequality $\sin^{-1}\left(\sin\left(\frac{2x^2+4}{1+x^2}\right)\right) < \pi -3$?

Question

$$\sin^{-1}\left(\sin\left(\frac{2x^2+4}{1+x^2}\right)\right) < \pi -3$$

I don't know where to start here because of the inverse sine function

I know $\sin^{-1}(\sin x) = x$ if x lies between -1 and 1, but I don't know how to apply it here

Answer 1

$$\sin^{-1}\left(\sin\left(\frac{2x^2+4}{1+x^2}\right)\right) < \sin^{-1}(\sin(\pi -3))$$

Inverse sine is monotonous function so:

$$\sin\left(\frac{2x^2+4}{1+x^2}\right) < \sin(\pi -3)$$

This relation is of the form:

$$\sin y\lt\sin\alpha \quad(0<\alpha<\frac{\pi}{2})$$

..and can be solved in a fairly simple way.

HINT: The last inequality is equivalent to:

$$(-\pi-\alpha)+2k\pi\lt y \lt \alpha + 2k\pi \quad (k\in Z)$$

$$-2\pi+3+2k\pi\lt \frac{2x^2+4}{1+x^2} \lt \pi-3+2k\pi \quad (k\in Z)$$

$$-2\pi+3+2k\pi\lt 2 + \frac{2}{1+x^2} \lt \pi-3+2k\pi \quad (k\in Z)$$

$$-2\pi+1+2k\pi\lt \frac{2}{1+x^2} \lt \pi-5+2k\pi \quad (k\in Z)$$

$$\frac{-2\pi+1}{2}+k\pi\lt \frac{1}{1+x^2} \lt \frac{\pi-5}{2}+k\pi \quad (k\in Z)$$

You can take it from here...

Answer 2

Although I was the one to ask the question, I came up with a solution after thinking for a while:

We are given that:

$$\sin^{-1}\bigg(\sin\left(\frac{2x^2+4}{1+x^2}\right)\bigg) < \pi -3$$

We can re-write this as:

$$\sin^{-1}\bigg(\sin\left(\frac{2(1+ x^2) + 2}{1+x^2}\right)\bigg) < \pi -3$$

or

$$\sin^{-1}\bigg(\sin\left(2+ \frac{2}{1+x^2}\right)\bigg) < \pi -3$$

Now, $\forall \ x \in \mathcal R $, $0 <\displaystyle{\frac{2}{1+x^2}} \leq 2$, so $2 <2+\displaystyle{\frac{2}{1+x^2}} \leq 4$

So the whole thing can be re-written as: $ \displaystyle\sin^{-1}(\sin\theta) < \pi -3$ where $\theta$ is $2+\displaystyle{\frac{2}{1+x^2}} $

We also found out that $2 < \theta \leq 4$, but $ \frac{\pi}2 <2< \theta \leq 4 < \frac{3\pi}2$, so $\theta \in (\frac{\pi}2, \frac{3\pi}2 )$

Since $\theta$ lies in this interval, we can write $\sin^{-1}(\sin\theta)$ as $ \pi - \theta$, which can be seen from the graph below of $\sin^{-1}(\sin x)$

So, $$ \displaystyle\sin^{-1}(\sin\theta) < \pi -3$$

$$ \displaystyle \implies \pi - \theta < \pi -3\implies \theta > 3$$

combining this result with $ 2 < \theta \leq 4$, we get $ 3 < \theta \leq 4$

Substituting $\theta $ in terms of $x$,

$$ 3 < 2+\displaystyle{\frac{2}{1+x^2}} \leq 4$$

$$ \implies 1 <\displaystyle{\frac{2}{1+x^2}} \leq 2$$

Since $1+x^2$ is always positive, we can just multiply it,

$$ \implies 1+x^2 <2 \leq 2 + 2x^2 $$

this gives us $ x^2 <1 $ which means $x \in (-1,1)$