INSTRUCTOR: Hi, everyone. Today we're going to look at solving trig equations using identities. The biggest hint when I look at an equation that I'm going to have to use an identity and I can't just use order of operations backwards or factor or something easier, the biggest hint is going to be that I have more than one trig function represented in my equation. 

So here, I notice that I have a secant squared and a tan squared. So my goal is going to be to get the equation entirely in terms of one trig function. The first identity that I'm going to want to use here, since I see a secant squared and a tan squared, I'm probably going to want to use one of the Pythagorean identities. The Pythagorean identity that I'm going to want to use here is that secant squared x is equal to 1 plus tan squared x. If you want some easy ways to help you memorize the identities or figure them out, if you want an easy way to either derive or memorize the identities that we're going to be using today, please take a look at the video linked in the comments below for deriving trig identities. 

So since I know that secant squared is equal to 1 plus tan squared, I'm going to replace that secant squared with 1 plus tan squared. And then I'm just going to carry down the rest of my equation. From here, I'm going to continue solving. 

So first, I'm just going to get rid of these parentheses by distributing the two, carry down the rest of my equation. I can combine the 2 tan squared and the 1 tan squared to become 3 tan squared. And positive 2 minus 3 gives me minus 1. Now I can solve this doing order of operations backwards. So I'm going to add the 1, divide by 3, take the square root, and I get plus or minus the square root of 1 over 3. This is equal to plus or minus 1 over the square root of 3. And this is not an exact value that I know, so I'm going to rationalize this by multiplying by rad 3 over rad 3. 

When I do that, I get tan x is equal to plus or minus rad 3 over 3. That is one of my exact values. The reference angle for tan x equal to rad 3 over 3 is pi over 6. I know that since my answer is plus or minus that I'm going to have answers in all four quadrants. And I also know that in order to figure out the angles based on the reference angle in each of the other three quadrants that the numerator depends on the denominator. 

So in the second quadrant, the numerator is going to be minus 1 from the denominator. In the third quadrant, it's going to be plus 1. And in the fourth quadrant, it's going to be times 2 minus 1. So my four answers for x are going to be, in the first quadrant, pi over 6, just the reference angle. In that second quadrant, minus 1 from the denominator, so 5 pi over 6. In the third quadrant, plus 1 from the denominator, 7 pi over 6. And in the fourth quadrant, times 2 minus 1, so 11 pi over 6. 

Number 2. I know that I'm going to need an identity here because I have a sine squared x and a cosine x. So I have two different trig functions. Again, my goal is going to be to get this entire equation in terms of one trig function. I can't replace anything for cosine x, but when I see sine squared x, I'm immediately thinking the Pythagorean identity that says sine squared x plus cosine squared x is equal to 1. 

If I subtract the cosine squared over, I get sine squared is equal to 1 minus cosine squared. In my original equation, I'm going to replace the sine squared x with what it's equal to, the 1 minus cosine squared. So I have 2 times, in parentheses, 1 minus cosine squared is equal to 2 plus cosine x. I distribute the 2. I get 2 minus 2 cosine squared x. And that's going to be equal to 2 plus cosine x. 

From here, since I have a cosine squared x and a cosine x, I have variables on both sides of the equation, I'm going to move everything over so I can set the equation equal to 0. 2 minus 2, that just cancels out and becomes 0. So I'm going to be left with 2 cosine squared x plus cosine x equal to 0. From here, I see that I have a GCF of cosine x. So if I factor out a cosine x, I'm left with 2 cosine x plus 1 and then equal to 0. 

If I tee it up and solve, on the left side, I have cosine x equals 0. And I know that cosine of x is equal to 0 at pi over 2 and 3 pi over 2. On the right side, I have 2 cosine x plus 1 is equal to 0. So if I subtract 1 and divide by 2, I get cosine x is equal to negative 1/2. The reference angle for cosine x equal to negative 1/2 is pi over 3. 

And I'm going to draw my little chart for ASTC. I'm looking for where cosine is negative. Cosine is negative in the second quadrant and in the third quadrant. So my two solutions are going to be, OK, in the second quadrant, I'm minus 1, so 2 pi over 3. And in the third quadrant, I'm plus 1, so 4 pi over 3. So I have four solutions for number 2-- pi over 2, 3 pi over 2, 2 pi over 3, and 4 pi over 3. 

Number 3, I have 8 sine theta is equal to sine of 2 theta. Now, I have the same trig function on both sides. My hint here that I'm going to need an identity is this sine of 2 theta. That's jumping out at me as one of my double angle identities. I know that sine of 2 theta is equal to 2 sine theta cosine theta. 

So what I'm going to do is first replace the sine of 2 theta with 2 sine theta cosine theta. From here, I'm going to get this thing equal to 0 so that hopefully I can start factoring something out. So if I subtract the 2 sine theta cosine theta over, I can now see that I have a GCF of 2 sine theta. So if I factor out a 2 sine theta, I'm left with 4 minus cosine theta. 

If I tee this up and solve, over here I have 2 sine theta is equal to 0. If I divide by 2, I have sine theta equals 0. And I know that sine is equal to 0 at 0 pi and 2 pi. On the right side, if I add the cosine over, I get 4 is equal to cosine theta. And I know that cosine theta has to equal something between negative 1 and positive 1. And since 4 is outside of that interval, this side has no solution. So my three answers for number 3 are 0, pi, and 2 pi. 

And number 4. I have two trig functions showing up here. And I'm also noticing that I have this cosine of 2 theta. So again, that's another double angle identity. I also know that cosine of 2 theta has three different forms. The form that's going to be the most useful to me in this scenario is going to be the one that only has sine in it so that this equation becomes entirely in terms of sine theta. 

I know that cosine of 2 theta is equal to 1 minus 2 sine squared theta. So if I sub that in, I have 1 minus 2 sine squared theta minus sine theta plus 2 is equal to 0. If I move stuff around here a little bit, I'm going to end up with negative 2 sine squared minus sine plus 3. I don't want to have to factor this with a negative leading coefficient, so I'm going to divide by negative 1 and make this 2 sine squared plus sine minus 3 is equal to 0. 

From here, I'm going to let sine theta equal x so that this thing is a little bit nicer to factor. So I'm going to write this now as 2x squared plus x minus 3. I'm looking now for two numbers that add to positive 1 but multiply to negative 6. Those numbers are going to be positive 3 and negative 2. I'm going to split up this middle term, the plus x, into a positive 3x and a minus 2x. 

So I'm going to write this as 2x squared minus 2x plus 3x minus 3 is equal to 0. And now I do factor by grouping. I pull out a 2x in the first two terms. I'm left with x minus 1. I pull out a 3 in the second two terms. I'm left with x minus 1. I can now factor in x minus 1 out, and I'm left with 2x plus 3. From here, if I substitute sine theta back in, I'm left with sine theta minus 1 times 2 sine theta plus 3. 

Tee this up and solve. On the left side, I get sine theta equal to 1. I know that sine theta is equal to 1 when theta is equal to pi over 2. On the right side, I end up with sine theta is equal to negative 3 over 2. I know that that's smaller than negative 1, so I have a no solution over there. My only solution to this problem is theta equals pi over 2. 

Number 5. I see right away that I have two different trig functions in here, so I'm first going to want to rewrite so that this is in terms of sine. And I'm also going to distribute the 2 as I go. So 2 times sine theta is going to give me 2 sine theta. I know that cosecant theta is equal to 1 over sine theta. So I end up with 2 over sine theta is equal to 5. 

My equation is entirely in terms of sine, but this denominator is kind of a pain. So what I'm going to do is multiply this entire equation by sine theta to clear that denominator. So when I do that, 2 sine theta times sine theta is going to give me 2 sine squared theta. Sine theta times 2 over sine theta, the sine thetas cancel. I'm left with just plus 2. And 5 times sine theta gives me 5 sine theta. 

Since I have a sine squared and a sine, I know that I'm going to want to factor this thing. So I'm going to subtract the 5 sine theta over. And now I'm going to start factoring. So first I'm going to let sine theta equal x. And I'm going to rewrite this as 2x squared minus 5x plus 2. I know that I'm now looking for two numbers that add to negative 5 but multiply to positive 4. So that's going to be negative 4 and negative 1. 

I'm going to split up this middle term so that it becomes negative 4x and negative x. Copy down the rest of my equation. And now I do factor by grouping. In my first two terms, I can pull out a 2x. I'm left with x minus 2. In the second equation, I'm going to pull out a negative 1-- or sorry, in the second two terms, I'm going to pull out a negative 1. And I'm left with x minus 2. Now I can factor out an x minus 2, and I'm left with 2x minus 1. 

I'm now going to sub the sine theta back in for x so that I have sine theta minus 2 times 2 sine theta minus 1. On the left side, I get sine theta is equal to 2. I know that's a no solution. On the right side, I get sine theta is equal to positive 1/2. Since sine is positive, I'm going to have answers in the first quadrant and in the second quadrant. 

My reference angle for sine theta equals 1/2 is pi over 6. So in the first quadrant, that's just going to be the reference angle itself. And in the second quadrant, it's going to be minus 1 from the denominator, so 5 pi over 6. So I have two solutions here. That's it for solving trig equations using identities. If you have any questions, feel free to leave them in the comments below. Have a great day.