Question

Find the solutions for each equation on [0,2pi).
1. 2sin^2Ө=1-sinӨ
2. 2tanӨ-sec^2Ө=0
3. sin2Ө+sinӨ=0

Answer 1

Well let's start with the first one. This is a quadratic equation, but instead of x we have sine. So we want to factor this just liek a quadratic, which means we set everything equal to 0 and factor:
\[2\sin ^{2}\theta+\sin \theta-1=0\]Now if you were to pretend that sin(x) was just x, would you know how to factor that quadratic?

Answer 2

we are pretending the sinӨ is x?

Answer 3

Correct. Because this is done EXACTLY like if we had
\[2x ^{2}+x -1 = 0\]we factor it the same way, its just its sinx instead of x.

Answer 4

(2x-1)(x+1)=0 got it whats next?

Answer 5

Now we just need to replace x with sin x and set both factors equal to 0. So we have this:
\[2\sin \theta-1 = 0\]and
\[\sin \theta +1 = 0\]
Now we first solve for sin(theta) then solve for the actual theta.

Answer 6

and you do that for both? So
sinӨ=1/2
and
sinӨ= -1

Answer 7

Correct. Now do you know the values of the unit circle or have a unit circle chart handy?

Answer 8

yes how do I used that?

Answer 9

Right. Well, the unit circle chart will tell you where sin = 1/2 and where sin = -1. If you have a way to look it up then you can find which angles by looking at the chart.

Answer 10

and the angles are my answer?

Answer 11

Correct. You will have two angles that equal 1/2 and one angle that = -1. So you will have 3 answers.

Answer 12

okay thanks may need help with other two but don't know quite yet. :)

Answer 13

Did you find the correct angles?

Answer 14

When looking at the circle is the 1/2 and -1 in the "x" or "y" part of the coordinate

Answer 15

Sin refers to the y-coordinate.

Answer 16

30, 270, and 350 then

Answer 17

150 instead of 350.

Answer 18

okay

Answer 19

Now the second one requires us to use an identity. Do you know of this identity?
\[\tan ^{2}\theta+1 = \sec ^{2}\theta \]?

Answer 20

uhm yes I don't remember how to use it tho?

Answer 21

Well our second problem is also a quadratic. The problem is we have a tangent and a secant. In order to do this problem, they must both be the same trig function. So we use this identity to change secant into tangent. Now we have:
\[2\tan \theta -\sec ^{2}\theta=0\]But look what the identity above says. It says secant squared is the same as tangent squared + 1. So we have to substitute that.

Answer 22

okay

Answer 23

Now just make sure to subsitute it properly, not making sign mistakes:

\[2\tan \theta -(\tan ^{2}\theta +1)=0\]Now would you know how to solve that in the same way we did with the first problem?

Answer 24

yes I would thanks

Answer 25

Alright, cool. So we would end up with
\[\tan \theta = 1\]Now to solve this one, we need to know that:
\[\tan \theta = \frac{ \sin \theta }{ \cos \theta }\]So when we check our unit circle, we want to know when:
\[\frac{ \sin \theta }{ \cos \theta }=1 \]There will be two answers :3

Answer 26

If you got the answer to this one we can go to the last one.

Answer 27

no sorry I didn't get that one yet

Answer 28

Ah, okay. Well, we need to find angles where if we divide sin and cos we get 1. This means that the values for sin and cosine must be the same at the angles we need. There are two spots where this is true.

Answer 29

90 and 360

Answer 30

Well, at 90 degrees cosine is 0 and at 360 sin is 0, so theres no way that they can divide and get you 1. In fact, tangent is completely undefined at 90.

Answer 31

so am I looking for sin and cosine at 1 or tangent?

Answer 32

You want a tangent that is 1. But to get a tangent that is 1, you need to find out where sin and cosine are the exact same value. Since tangent = sin/cos, the only way two numbers can divide and equal 1 is if they are the exact same number.

Answer 33

do you have a unit circle handy tat you could help me? I'm looking at this one and just cant figure it out

Answer 34

Im sure we can google and find one. I just have the values memorized personally. Um...lemme look.

Answer 35

okay thanks:)

Answer 36

http://www.embeddedmath.com/downloads/files/unitcircle/unitcircle-letter.pdf

Answer 37

There are two spots where sine and cosine are the exact same. And remember that by exact same, that means the same sign as well.

Answer 38

wait im writing out my work and can we go back to where we switched sec^2 theta to tan^2+1 what happens after I do that step im sorry

Answer 39

Ah, okay. Well, we had:
\[2\tan \theta - (\tan ^{2}\theta + 1) = 0\]Set everything equal to 0 and make sure the tan^2 term is positive: So we'll add the quantity of tan^2 + 1 to the other side of the equation and then move 2tan to the other side, too: So we end up with:
\[\tan ^{2}\theta - 2\tan \theta +1 = 0\] Now we just factor this liek a quadratic, same way we did in the first problem.

Answer 40

and could u write out the equations so I can see if I'm correct?

Answer 41

Factors of 1 that add up to negative 2 = (-1)(-1). Therefore we factor this into:
\[(\tan \theta -1)^{2}=0\]The two factors are the same,so we set one of them equal to 0:
\[\tan \theta =1 \]Now we find what angles havea tangent value of 1. This is done by checking our unit circle and seeing where sine and cosine are the exact same values. THis occurs at two angles.

Answer 42

and both sine and cosine are the y coordinate?

Answer 43

Cosine is the x-coordinate and sine is the y-coordinate.

Answer 44

is it 60 and 30?

Answer 45

Neither. at 30 degrees, sin is 1/2 and cosine is sqrt(3)/2. At 60 degrees it is flip-flopped. sine is sqrt(3)/2 and cosine is 1/2. We want sine and cosine to be the exact same values. That link I posted of the unti circle should show you.

Answer 46

45 and 225?

Answer 47

There you go xD

Answer 48

yay!! can we do the last one? :)

Answer 49

Yeah, sure. Now this one will be a little bit different. We do have everything sine, but we have different angles. One is 2theta, the other is just theta. We cannot have that, these must be the same angle. So this means we have to use the double-angle formula on the sin2(theta). Would you happen to know it?

Answer 50

nope LOL but if you do just show me the work I can usually teach myself

Answer 51

Well, the double-angle formula for sin2(theta) is:
\[2\sin \theta \cos \theta \]
So this means we rewrite what we have as:
\[2\sin \theta \cos \theta + \sin \theta = 0\]So now we just have to factor this the best we can, which isn't much, but you see how to do it?

Answer 52

yes but now how would you factor that mess LOL?

Answer 53

Only one thing you can do and thats factor out a sin xD

Answer 54

so sin(1/cosӨ)=0
and
sinӨ=0?

Answer 55

Nope: Like this:
\[\sin \theta(2\cos \theta +1)=0 \]can you see how?

Answer 56

so im only gonna have one function?

Answer 57

Nope, you have 2. sin(theta) counts as a factor, too. You set both factors to 0 like this:
\[\sin \theta = 0\]
\[2\cos \theta + 1 = 0\]

Answer 58

then so Ill end up with 0 and -1/2 right?

Answer 59

0 being sine and -1/2 being cosine

Answer 60

Thats what theyll be equal to. Now you need to solve for theta in both equations.

Answer 61

by the unit circle right? how many angles should I get

Answer 62

4 : )
2 for each.

Answer 63

240, 120, 180, and 360?

Answer 64

Sounds good xDD

Answer 65

YES!!! Thank soooo much:) I really appreciate it

Answer 66

yep, no problem :3