00:01
My approach to this problem is that cosecant of 3 theta is the same thing as 1 over sine of 3 theta.
00:14
So i just rewrite this other thing.
00:16
Now, just multiplying over, what i'm going to end up with is sine squared of 3 theta.
00:24
But then i can also divide that 7, because it's a multiplication problem to the left side.
00:28
And then when i square root, i just have to make sure i have the positive and negative version.
00:34
So let me do it this way.
00:41
3 theta will equal the positive version of the inverse sine of 1 7th.
00:48
Just make sure that you're in the radian mode.
00:51
So inverse, oops, i forgot to write the square root.
00:56
Inverse sine of the square root of 1 7th gives me an answer of 0 .3876.
01:07
Hopefully that's enough decimals.
01:10
But also, i have to do the negative version of that, so positive and a negative version.
01:22
That i get negative 0 .3876.
01:26
So what's going to happen, though, is each one of these answers can repeat every pi radians in the first and third quadrant, and then in the second and fourth quadrant, 0 .3876 plus pi.
01:42
And what i need to do is divide each of those by 3.
01:45
So if i divide by 3, i get theta is about 0 .129...