00:01
Okay, we're going to find all the solutions to the equation given on the interval from zero to two pi.
00:08
So the first thing we want to do is we want to move everything to the same side of the equation.
00:13
If we actually divided out that cosecant x, we'd be losing some solutions.
00:18
So let's go ahead and subtract that to co -secon x.
00:26
Okay, so now that we are equal to zero, we can consider factoring.
00:32
Since both terms have that co -seekin x, we can factor it out.
00:37
What we're left with is secant x minus 2.
00:41
So we can use our zero product rule and set each one equal to zero separately.
00:47
So we have co -seken x equals zero, which means one over sine x would equal zero...