How to solve this inequality?
$$\arccos(x) \gt \arccos(x^2)$$
I've brushed up on arccos and it's basically the inverse function of the cosine and returns a degree value(or radians). Also, the domain of this function is -1, 1 and the range from 0 to pi.
$\endgroup$2 Answers
$\begingroup$Hint: $\cos$ is strictly decreasing on $[0,\pi]$, so $\arccos$ is strictly decreasing on $[-1,1]$. Hence $\arccos(x)>\arccos(x^2)$ is equivalent to $x<x^2$.
$\endgroup$ 2 $\begingroup$As the function $\arccos$ is monotonous decreasing,
$$\arccos(a)<\arccos(b)\implies a>b.$$
The given equation turns to $$x<x^2$$ $$0<x(x-1),$$ $$x<0\lor x>1.$$
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