How ro solve this? $arccosx>arccosx^2$ .

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Answer 1 · 2017-04-28T09:01:09

$-1 < x < 0$

Using

$Sin(a-b)=sina cosb-cosa sinb$

we have

$sin(arccosx-arccosx^2)=x^2 sqrt[1 - x^2] - x sqrt[1 - x^4] > 0$

or

$x sqrt[1 - x^2] (x - sqrt[1 + x^2]) > 0$

now assuming $sqrt[1 - x^2] > 0$ for $abs x < 1$ we have

$x(x - sqrt[1 + x^2]) > 0$

but

$sqrt[1 + x^2] > x$ so we need

$-1 < x < 0$

How ro solve this? arccosx>arccosx^2arccosx>arccosx^2.