00:01
Problem for being asked to solve the given equation for values of theta between 0 and 2 pi.
00:06
So notice that your answers need to be in radiant mode.
00:09
So the first thing i notice, we have two different trig functions.
00:12
We have cosine and we have sine.
00:14
So the first thing we're going to want to try and do is rewrite this equation so that way they both have the same trig function.
00:21
Well, in order to do this, i'm going to use the pythagrin identity that says the sine square the theta plus the cosine square the theta is equal to 1.
00:30
Because i can solve for the cosine square the theta by subtracting the sine square the theta from both sides.
00:37
So that will leave us with the cosine square the theta equal to 1 minus sine square the theta.
00:47
So now, in place of cosine squared theta in our problem, i'm going to substitute in 1 minus sine square the theta.
00:53
So we're going to have 2 times 1 minus sine square the theta, and this is equal to 2 plus the sine of theta.
01:01
So again, now we only have the sine function.
01:04
Next thing i'm going to do is get rid of the parentheses by distributing, well, 2 times 1 is 2, and 2 times negative sine square the theta, and the right -hand side stays the same.
01:18
So now the next thing i'm going to do is set my equation equal to 0.
01:21
So i'm going to move the terms on the left over to the right.
01:24
So first, i'm going to subtract 2, but when i do that, both of my 2 is cancel.
01:29
At the same time, i can add 2 sine square the theta to both sides of the equation, because the terms on the left cancel.
01:38
So i'll have zero equals, and if i put this in standard form, i'm going to have two sine square the theta plus the sine of theta.
01:47
Next, what i'm going to do is i'm going to solve this by factoring.
01:51
I'm going to take out a gcf of sine of theta...