The n is because sin is periodic every 2pi radians, this is my interpretation of the answer but a markscheme would be nice to check if this is what your question wants.
Not 100% sure, but arcsin(0) = 0 +(2pi)n where n is a integer.
0 + 2(pi)n = x + pi/6
x = 2pi(n) - pi/6
The n is because sin is periodic every 2pi radians, this is my interpretation of the answer but a markscheme would be nice to check if this is what your question wants.
Yeah this isn't true. If this was the case then arcsin would not be a function. It has a restricted range so that it is a function, namely [−2π,2π].
If you're saying what you've highlighted in bold, what you mean is that any number with sine 0 is 0+2nπ (for some
QUOTE=16characterlimit;65785569]Not 100% sure, but arcsin(0) = 0 +(2pi)n where n is a integer.
0 + 2(pi)n = x + pi/6
x = 2pi(n) - pi/6
The n is because sin is periodic every 2pi radians, this is my interpretation of the answer but a markscheme would be nice to check if this is what your question wants.
Yes, I got x = 2pi(n) -pi/6 but the mark scheme says x = pi(n) - pi/6, I'm not sure if both are accepted?
I think I could maybe work it out now by considering x + pi/6 as a translation to the left by pi/6. If you look at the graph the answer is computed for every pi(n) - pi/6, but I don't understand how to get the answer if I'm not doing it like that.