00:01
Okay, when solving this trig equation that contains a multiple argument, we first want to isolate the trig function.
00:08
We would do that by subtracting one from both sides of the equation, and then dividing both sides of the equation by the square root of three.
00:21
Okay, and then we're going to solve this over all real numbers or the general solution, and then also over zero to two pi.
00:29
To do so, we want to think about first, where is tangent negative 1 over root 3? regardless of the 3 theta or this horizontal compression.
00:39
So we're essentially ignoring this first.
00:41
And tangent is negative 1 over root 3 at 5 pi over 6 and 11 pi over 6.
00:47
We would write that general solution as 5 pi over 6 plus pi k.
00:54
We want to set this equal to 3 theta so that we can now apply the horizontal compression by multiplying by 1 third or dividing by 3.
01:05
This leaves us with 5 pi over 18 plus pi divided by 3 times k.
01:16
And then this is considered our solution over all real numbers.
01:20
When we're ready to solve over 0 to 2 pi, we want all of the solutions that that general solution provides for us, starting with k equals 0...