solve

The solution given by eureka123 is correct as far as principle domain is concerned....

But in general,

the solution is 2n(pi)≤x≤(2n+1)(pi) ; n=integer

This can be arrived at by seeing the values of sinx about the four quardrants

[339]

You guys really need to brush up your fundas.

Please check up what are the domains and ranges of the inverse trigonometric functions.

sin^-1x ≥0
Now x=[-1,1] for inverse function
now range of sin^-1x is [-∩/2,∩/2]
So sin^-1x≥0 when x = [0,1]