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#