solve abs(x^(2)-1+sinx)=abs(x^(2)-1)+abs...

solve `abs(x^(2)-1+sinx)=abs(x^(2)-1)+abs(sinx)`, where `x in [-2pi,2pi]`.

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The correct Answer is:

`x in [-2pi,-pi] cup [-1,0] cup [1,pi]`

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