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...

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].