Trigonometric inequalities: when to reverse sign I have the inequality 
$$\cos^{-1}{x^2 \over 2x-1} \ge {\pi\over 2}  $$
Now, by multiplying both sides by cos, I get.
$$ {x^2 \over 2x-1} \ge 0$$
However, I SHOULD be getting 
$$ -1\le {x^2\over2x-1} \le 0$$
And this raises three questions


*

*Where does that $-1$ come from?

*Why were the signs reversed, and how do I know when to do it? 

*Where do I go from here?

 A: You're not multiplying both sides by $\cos$; rather, you are taking the cosine of both sides.
The reason the direction of the inequality gets inverted, so that it says "$\ge$" rather than "$\le$", is that the cosine decreases as its argument increases from $0$ to $\pi$.
And cosines of real numbers are always in the interval $[-1,1]$, so that's why it's $\ge-1$.
The numerator is positive except when $x=0$, so the fraction as a whole is positive or negative according as the denominator is positive or negative.  You have
$$
2x-1 = 2\left(x-\frac 1 2\right)
$$
and that is positive if $x\ge\dfrac12$ and negative if $x\le\dfrac 1 2$.  It approaches $+\infty$ as $x$ decreases to $1/2$ and $-\infty$ as $x$ increases to $1/2$.
For the inequality $-1\le\dfrac{x^2}{2x-1}$, adding $1$ to both sides yields
$$
0\le1 + \frac{x^2}{2x-1} = \frac{x^2+2x-1}{2x-1} = \frac{(x^2+2x+1)-2}{2x-1} = \frac{(x+1)^2-2}{2x-1}
$$
$$
=\frac{(x+1-\sqrt2)(x+1+\sqrt2)}{2\left(x-\frac12\right)}.
$$
So examine the changes in sign at $1/2$ and at $1\pm\sqrt2$.
