Question

Solve the following equations algebraically to determine exact solutions over the interval [0.2pi]
a) sin 2x +sin 4x = 0
b) cos 2x +sin x =0

Answer 1

Part (b) first

Answer 2

cos (2x) = 1 - 2 sin^2 (x)..thats one of the formulas on your formula sheet

Answer 3

so, 1 - 2 sin^2 (x) + sin x = 0

Answer 4

Or, 2 sin^2 (x) - sin x - 1 = 0

Answer 5

ok

Answer 6

Now factor, (2sinx + 1)(sinx - 1) = 0

Answer 7

Set each factor equal to zero,
the first factor (2 sin x + 1) = 0, so sin x = - 1/2, that occurs at 210 and 330 degrees, or - 7pi/6 and 11pi/6 radians.

Answer 8

The second factor, (sin x - 1) = 0 means sin x = 1, and that occurs only at 90 degrees, or pi/2 radians.

Answer 9

The second factor, (sin x - 1) = 0 means sin x = 1, and that occurs only at 90 degrees, or pi/2 radians.

Answer 10

Final answers to part(b) equation is pi/2, 7pi/6, and 11pi/6.

I think I typed -7pi/6 earlier, thats not right, I meant 7pi/6

Answer 11

We are now done with part (b). What are your questions in this part (b)?

Answer 12

When you went from 1-2sin^2x + sinx to 2sin^2x - sinx -1
How did you do that?

Answer 13

You can change the signs of everything...why do I want to change the signs? Because I want to factor something that starts with 2 sin^2 (x) and NOT - 2sin^2(x).

Answer 14

So you can change the signs as long as you change every single one.

Answer 15

Yes, take a look for a moment.

We had 1 - 2 sin^2(x) + sin x = 0
So, go ahead and bring everything from the left side to the right side, what do you get?
You get 2 sin^2(x) - sin x - 1 = 0
Agree?

Answer 16

Yes

Answer 17

So you can bring slowly everything from the left side to the right side, OR you can change all the signs, and leave the left side as it is (with the signs changed).

Answer 18

I just changed all the signs, and left the left side where it was, with just the signs changed.

Answer 19

Alright, I see that.

Answer 20

Do you also see, how I got 7pi/6 and 11pi/6 and pi/2 as the solutions? Thats critical.

Answer 21

We wanted, as stated in the directions, all angles x in the closed interval [0,2pi].

Answer 22

Yes. By using the unit circle, and plugging in the "coordinates" of sin (or cos)

Answer 23

ok. Part (b) was not simple. Part (a) is much trickier, so I left that for last. So I want to take a 2-minute break and will then return to show you part (a). I just need to get a drink.

Answer 24

Ok :)

Answer 25

Will be back in approx 2 min

Answer 26

Alright

Answer 27

I'm back. Ready when you are.

Answer 28

The equastion in part (a) is sin(2x) + sin (4x) = 0.

Answer 29

Ready?

Answer 30

Yep

Answer 31

OK. Now, we have a formula for the sin(2x) = 2 sin (x) cos(x).
We do NOT have a formula for sin (4x), although we can develop that formula, but I do not think your teacher would want you to take that approach. I will give you a solution which I feel your teacher would want you to present.

Answer 32

In our equation, let u= 2x.
Si if u = 2x, then 2u = 4x. Agree?

Answer 33

Yes

Answer 34

So our equation becomes sin (u) + sin (2u) = 0

Agree? I will wait for you.

Answer 35

?

Answer 36

I got it.

Answer 37

sin (2u) = 2 sin(u) cos (u)

Answer 38

so our equation is now sin (u) + 2 sin(u)cos(u) = 0

Answer 39

Factor out sin (u), we get
sin (u) [1 + 2 cos (u)] = 0

Answer 40

set each factor equal to zero,
so sin (u) = 0 AND 2 cos(u) = -1
Or, sin (u) = 0 AND cos (u) = -1/2

Answer 41

Agree so far?

Answer 42

Yes

Answer 43

Now, let's take the first factor and set equal to zero and solve.
sin (u) = 0, then u = 0, pi, and 2pi
BUT we want to solve for x, not u, so we substitute u = 2x
therefore, 2x = 0, 2x = pi, and 2x = 2pi
hence, x = 0, pi/2,pi, 3pi/2, 2pi

Answer 44

We take the second factor, cos (u) = -1/2, and u = 2pi/3, 4pi/3, etc.but u = 2x, so x = pi/3, 2pi/3, pi,4pi/3, and 5pi/3

Answer 45

Done. So there are many solutions. Now, look everything over and tell me what you dont follow, if anything.

Answer 46

There?

Answer 47

Tell me if I made any error, as I do all of this in my head.

Answer 48

How did you get 3pi/2 and 2pi for the first factor?

Answer 49

We want all solutions in the closed interval {0,2pi]. Look at the first 3 solutions there, o, pi/2, pi...so every pi/2 theres another solution..so i added pi/2 to pi to get 3pi/2, then i added another pi/2 to 3pi/2 to get 2pi

Answer 50

?

Answer 51

Ok

Answer 52

Go through each step and you'll see that each step makes sense. It is a tough problem. But the concept of letting u = 2x and 2u = 4x is critical here.

Answer 53

Feel free to get back to me if any step is not crystal clear.

Answer 54

Thank you! I'll definitely do that (:

Answer 55

No problem. Your welcome.