math inverse trignometric question..

k must be positive, so x should be positive and in the 1st or 2nd quadrant.. But because there's an inverse of the sin function which has a range -90<=sinx<=90, the only positive values of x that are allowed should be 0<=x<=90.. so the only possible value for x should be x=a.. but ms says its x=a and x=180-a... pretty sure I'm making a stupid mistake here without even realizing it, but i have no idea what that might be.

Youre sort of right. There is a single output/solution of arcsin(k) = alpha, so alpha is in quadrant 1 as its positive. However there are multiple solutions to
sin(x) = k
The first two being alpha and 180-alpha as the mark scheme says.

Its the "same" as saying x=sqrt(k) has a single solution/value (positve) but there are two solutions to
x^2 = k

I know there are multiple solutions, but isn't the only solution in range of -90<x<90 x=a? because that's the only solution for which inverse of sinx exists...... idek what I'm doing wrong here

Youre sort of right. There is a single output/solution of arcsin(k) = alpha, so alpha is in quadrant 1 as its positive. However there are multiple solutions to
sin(x) = k
The first two being alpha and 180-alpha as the mark scheme says.

Its the "same" as saying x=sqrt(k) has a single solution/value (positve) but there are two solutions to
x^2 = k

There is a single value of arcsin(k). arcsin() is the inverse of sin() only on the limited domain -90..90. However, sin() is not restricted to this domain so there are multiple/infinite number of solutions to
sin(x)=k
one of which is arcsin(k). Its the "same" as there are two solutions to x^2=k, but sqrt(k) is single valued.

wait.. I think i get it now.. there are multiple values of x for which sinx=k(as you said).despite one of those values being outside the range of arcsinx, it doesn't really matter, as the values which are outside of range (180-a) would still give the same value for k.. and this value of k gives a value which is within the range of the inverse sine function.. so both values x=a and x=180-a work... is that right?

Basically,
* multiple (infinite number of) values of x for which sin(x) = k
* only one of those values is in the range of arcsin(k)