# How to find the domain of y = arcsin(1+x/1-2x)?I know that x is an element of [-1,1] and x must not equal 1/2 but i am having a problem with the presentation.

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The domain of arcsin function is [-1;1] and the range is [-pi/2 ; pi/2].

The argument of the given arcsin function is (1+x)/(1-2x).

We'll impose that -1 =< (1+x)/(1-2x) =< 1

We'll solve double inequality:

-1 =< (1+x)/(1-2x)

(1+x)/(1-2x) + 1 >= 0

(1+x+1-2x)/(1-2x) >= 0

(2-x)/(1-2x) >= 0

For the ratio to be positive, both numerator and denominator has to be positive.

The ratio is positive if x belongs to [-1 ; 1/2).

x is not allowed to be equal with 1/2 since x=1/2 is the root of denominator and the denominator must no be zero.

We'll solve the other inequality:

(1+x)/(1-2x) =< 1

(1+x)/(1-2x) - 1=< 0

(1+x-1+2x)/(1-2x)=<0

3x/(1-2x)=<0

The fraction is negative if the numerator and denominator have different signs.

The fraction is negative if x belongs to the interval [-1;1/2)U(1/2;+1].

**The domain of the function is [-1 ; 1/2)U(1/2;+1].**