How ro solve this? $arccos x > {arccos x}^{2}$ $arccosx>arccosx^2$ .

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Answer 1 · 2017-04-28T09:01:09

$- 1 < x < 0$ $-1 < x < 0$

Using

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

we have

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

or

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

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

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

but

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

$- 1 < x < 0$ $-1 < x < 0$

How ro solve this? arccosx>arccosx2arccosx>arccosx^2.