00:01
We want to solve 6 sine of pi over 6 x is equal to 4 and find the four smallest positive solutions.
00:16
Let's first divide by 6 on both sides.
00:21
We get sine pi over 6 times x is 4 divided by 6, which is 2 3rds.
00:30
Then the inverse sine of 2 3rds will be equal to pi over 6 x.
00:40
Well, the inverse sine can give us multiple angles.
00:46
If we look at the unit circle, 2 3rds is here.
00:53
Then the inverse sine of 2 3rds is going to be the angle t that gives us this y coordinate in quadrant 1.
01:11
But also we're going to have the opposite angle in quadrant 2 that gives us that same y value.
01:21
So in q1 we have a solution, in q2 we have a solution, and then every time you go around a circle you're also going to get a solution.
01:35
So t1 t2 you're going to get t1 plus 2 pi k, where k is an integer greater than or equal to 0.
01:52
And here too you're going to get t2 from quadrant 2 plus 2 pi k for k 0 1 2 3 4 5 and going up to infinity.
02:10
So if we want the four smallest possible solutions, then we're going to find those two, those four sine of 2 3rds, then will give us the angle 0 .7297 in quadrant 1.
02:33
So that is the first one.
02:36
I'm not sure why i called these angles t, but anyway...