Question

sin4x= -2sin2x

Answer 1

Sine Double Angle Formula:
\[\Large \color{royalblue}{\sin 2x\quad=\quad 2\sin x \cos x}\]We're actually going to use this to turn our sin4x into a sin2x, k?\[\Large \sin4x\quad=\quad \sin2(2x)\quad=\quad 2\sin(2x)\cos(2x)\]Understand what I did there? :o

Answer 2

I understand that part, yes.

Answer 3

so 2sin(2x)cos(2x)= -2sin2x

Answer 4

\[\Large 2\sin2x \cos2x\quad=\quad -2\sin2x\]Ok good good :) hmm let's see.

Answer 5

Let's add 2sin2x to each side

Answer 6

alright, I'm pretty much stuck from there

Answer 7

Doing that last step gives us:
\[\Large \color{#CC0033}{2\sin2x} \cos2x+\color{#CC0033}{2\sin2x}\quad=\quad0\]Confused how to proceed from there? :o

Answer 8

2sin2xcos2x+2sin2x=0
2sin2x (cos2x+1)=0

Answer 9

good :) looks like you're on the right track!

Answer 10

I don't know what to do with the cos2x

Answer 11

So to continue, we'll set each factor equal to zero:\[\Large 2\sin2x=0\qquad\qquad\qquad \cos2x+1=0\]And solve for our trig functions in each case.
They should end up giving us some special angles.

Answer 12

\[\Large \cos2x+1=0\qquad\qquad\to\qquad\qquad \cos2x=-1\]Confused where to go from here?

Answer 13

yes

Answer 14

Think about this a sec:\[\Large \cos\color{royalblue}{\theta}\quad=\quad -1, \qquad\qquad\qquad\qquad \color{royalblue}{\theta}\quad=\quad ?\]

Answer 15

Do you remember your special angle that corresponds to this? :)

Answer 16

so x=pi and x=0?

Answer 17

Did the question restrict us to a certain domain or no?
It doesn't look like it did.
So we'll have to include `all` solutions which includes all rotations that would give us the same results.

Answer 18

no, no restriction on the domain

Answer 19

So for the sine solution:\[\Large \sin\color{royalblue}{2x}\quad=\quad 0, \qquad\qquad\qquad\qquad \color{royalblue}{2x}\quad=\quad n \pi\]All `whole number` multiples of pi, when we take the sine of them, should give us zero.

Example:
sin2pi=0
sin3pi=0
sin0pi=0

That's why I wrote n pi, k? :o
But we need to continue a little further and solve or x, not 2x.

Answer 20

so x=pi/2

Answer 21

not pi

Answer 22

and 3pi/2?

Answer 23

Yes good! :) Dividing by 2 gives us our first set of solutions:\[\Large x\quad=\quad \frac{n \pi}{2}\]Don't forget about the n!

Answer 24

The n takes care of 3pi/2, and all other multiples of pi over 2

Answer 25

alright, thanks :)

Answer 26

We solved for 2x first using what we know about our reference angles,\[\Large 2x\quad=\quad n \pi\]Then we divided by 2 to solve for x,\[\Large x\quad=\quad \frac{n \pi}{2}\]

Answer 27

We have another set of solutions coming from the cosine though ! :O

Answer 28

you mean the 4sinxcosx?

Answer 29

From this part:\[\Large \cos\color{royalblue}{2x}=-1,\qquad\qquad\qquad\qquad \color{royalblue}{2x}\quad=\quad ?\]

Answer 30

I thought this was the pi/2 part?

Answer 31

Mmmm no :(
Get confused by all the stuff going on? D:

Answer 32

cosine gives us -1, attttt 3pi/2 right?

Answer 33

no?

Answer 34

sine does

Answer 35

Oh ty ty silly brain of mine :) lol
At pi*

Answer 36

right

Answer 37

\[\Large \cos\color{royalblue}{2x}=-1,\qquad\qquad\qquad\qquad \color{royalblue}{2x}\quad=\quad n \pi\]Oh so this is giving us the same set of solutions actually.
So yah it works out nicely I guess.
We don't have to worry about anything.
Just the first set of solutions that we came up with:\[\Large x\quad=\quad \frac{n \pi}{2}\]Where n is any positive or negative integer.

Answer 38

we didn't go over the solutions to 2sin2x, but that just turns out to be x=nπ right?

Answer 39

That was the one we did at first :o
Start by dividing each side by 2, giving us:\[\Large \sin\color{royalblue}{2x}\quad=\quad 0, \qquad\qquad\qquad\qquad \color{royalblue}{2x}\quad=\quad n \pi\]We get zero at all integer multiples of pi.

Answer 40

Here is a graph just in case you wanted to see it in action :)
https://www.desmos.com/calculator/lkbans6evz

The red line is the sin4x
the green line is -2sin2x

You can click on the points of intersection and see that they're all multiplies of pi/2.
Cool stuff! :)

Answer 41

No, I don't think we'd done that yet

Answer 42

anyway we're finished now, thank you!

Answer 43

np :x