# What is x if the function arcsin(3-2x) exists?

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The sine function can take values only in the interval [-1, 1] irrespective of what the value of x is.

This implies that y = arc sin (3 - 2x) exists only for -1 <= (3 - 2x) <= 1

-1 <= (3 - 2x)

=> 2x <= 4

=> x <=2

(3 - 2x) <= 1

=> -2x <= -2

=> x => 1

**x can take values that lie in the interval [1, 2]**

We'll impose the constraint for the arcsin(3-2x) to exist:

-1 =< 3-2x =< 1

We'l solve the left side inequality:

-1 =< 3 - 2x

-1 - 3 =< -2x

-4 =< -2x => 4 >= 2x => 4/2 >= x

x =< 2

We'll solve the right side inequality:

3-2x =< 1

-2x =< 1 - 3

-2x =< -2

x >= 2/2

x >= 1

**The values of x, for the function arcsin(3-2x) to exist, belong to the closed interval [1 ; 2].**