106sin-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]
3 Answers
1The 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]
341You guys really need to brush up your fundas.
Please check up what are the domains and ranges of the inverse trigonometric functions.
106