# Trigonometric inequation $\sin^{-1} \frac 2x > \cos^{-1} \frac 2x$

So I have this trigonometric inequality (and I need to solve for $x$): $$\sin^{-1} \frac 2x > \cos^{-1} \frac 2x.$$ If I apply sine and cosine to both sides of the equation, I get $$\cos \frac 2x > \sin \frac 2x.$$

Can I do that? Also, if I can, in fact, do it, then what should be done in order to proceed?

Also, the answer I should be getting is the range $[2, 2\sqrt {2}]\,\,.$

Note that $\sin^{-1}\frac{2}{x}=\cos^{-1}\frac{\sqrt{x^2-4}}{x}$, so that your inequality is $$\cos^{-1}\frac{\sqrt{x^2-4}}{x}>\cos^{-1}\frac{2}{x}$$ Now take the cosine of both sides.
look at the unit circle with the point $P = (x,y)$ and $Q = (y,x)$ where we limit $1/\sqrt 2 \le x \le 1$ and $0 \le y \le 1/\sqrt 2.$ the point $P$ is on the unit circle between the $x$-axis and the diagonal line $y = x.$ call the point $A = (1,0), Q$ is the image of $P$ on the mirror $y = x.$
by definition $$\sin^{-1} y = arc(AP) \le arc(AQ) = \cos^{-1}y \text{ for } 1/\sqrt 2 \le y \le 1.$$
put $x = \dfrac{1}{y}$ to get the inequality you want.
$$sin^{-1} \frac 2x > cos^{-1} \frac 2x => sin^{-1} \frac 2x >\frac\pi4$$ Now looking at the graph of sine inverse, we get $$\frac 1{\sqrt {2}} < \frac 2x < 1 => 2 < x < 2\sqrt {2}$$