Find the at which value does the 1/(1-6sinx) is discontinous

[kougou]

Homework Statement

so, find the at which value does the 1/(1-6sinx) is discontinous

Homework Equations

The Attempt at a Solution

so, it's discontinous when the bottom is equal to zero, which is 1-6sinx=0

solve for x, which give arcsin(1/6).

and now, the correct answer is x is not = arcsin(1/6)+2pi n
and x is not = pi-arcsin(1/6)+2pi n

the bold part is where i don't understand. what? the arcsin has a period of 2pi? did we already restrict it to -pi/2 to pi/2?
what? how come the answer lays on the second quadrant? shouldn't it be lay either in first or the fourth quadraint?

what? how come the we ignore the third quadrant?

don't get it. need help

 

[Dick]

1-6*sin(arcsin(1/6)+2*pi*n) is equal to zero isn't it? sin is periodic with period 2pi, arcsin doesn't have to be.

 

[kougou]

1-6*sin(arcsin(1/6)+2*pi*n) is equal to zero isn't it? sin is periodic with period 2pi, arcsin doesn't have to be.

Hello

yes, sin is periodic, but why we add 2pi n at the end, and why we subtract that from pi. I don't get it. is there a way to derive this?

 

[Dick]

Hello

yes, sin is periodic, but why we add 2pi n at the end, and why we subtract that from pi. I don't get it. is there a way to derive this?

Look at a graph of sin(x) on [-pi,pi]. There are two values of x where sin(x)=1/6, yes? One is arcsin(1/6). The other is pi-arcsin(1/6). You can read those off the graph. Now you can add 2*pi*n to either one since sin is periodic.