Question

???

Answer 1

use line 1 to re-write 2sin(x)cos(x) in terms of sin only

Answer 2

hint: try replacing the alpha's with x's

sin(2x) = 2sin(x)cos(x)

Answer 3

wait so :
sin(2x) = 2sin(x)cos(x)

Answer 4

yes

now going back to the original equation

how can you rewrite 2sin(x)cos(x) in terms of sin only?

Answer 5

sin(2x)

Answer 6

good

then for part I our sol'n would be sin(2x) - sin(2x)cos(2x) = 0

for part II: try factoring out sin(2x)

Answer 7

4sin^3 sin(x) cos(x)

Answer 8

hm. that's not quite what I meant.

try to re-write sin(2x) - sin(2x)cos(2x)

so you have sin(2x) factored out like this:

sin(2x)[_____ - ______]

and fill in the blanks with the appropriate values after sin(2x) is factored out

Answer 9

pi/2

Answer 10

we aren't solving for x yet, just factoring

Answer 11

like, if we have something like

x - xy

we can factor out x to get

x(1-y)

can you try applying this same logic to

sin(2x) - sin(2x)cos(2x)?

Answer 12

it's basically the opposite of the distributive property

Answer 13

sin(2x)(- sin(2x)cos(2x))

Answer 14

hm, getting there but not quite

to make this a little easier you can divide each term by sin(2x)

so

sin(2x)/sin(2x) = ?

sin(2x)cos(2x)/sin(2x) = ?

Answer 15

cos(2x)

Answer 16

good but don't forget about the first expression

sin(2x)/sin(2x) = 1 since anything divided by itself is 1

Answer 17

so putting that together we get

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

sin(2x)[1-cos(2x)]

Answer 18

sin(2x)[1-cos(2x)] = 0 to be more specific

Answer 19

then for part III you have a case where

A*B = 0, so either A = 0 or B = 0, or both

applying this logic to our problem

sin(2x) = 0
[1-cos(2x)] = 0

try solving both these scenarios for x

Answer 20

this is mostly a matter of knowing your unit circle

let's start with sin(2x) = 0 b/c that's a little simpler

what theta value(s) have a sin value = 0?

Answer 21

0

Answer 22

ugh

Answer 23

good but there's one more check the other side of the UC

Answer 24

0 and pi

Answer 25

awesome

since we have sin(2x) not just sin(x) we have to divide those by 2

so 0 and pi/2

now, it asks for the general solution which means must account for equivalent angles

Answer 26

to account for this we can add 2pi*n to these values to get

0 + 2pi*n
and

pi/2 + 2pi*n

Answer 27

so this is part II

Answer 28

part II is just asking for the factored form so sin(2x)[1-cos(2x)] = 0

Answer 29

part III is getting into the solving part

so just for the sin values we have 0 + 2*pi*n and pi/2 + 2*pi*n for any integer n

then we just repeat this process for the cos values

Answer 30

hm

Answer 31

so just using cos 2x

Answer 32

so just what we have?

Answer 33

actually nvm it's alright so far

anyway the other part of the eqn is

[1-cos(2x)] = 0

so solving for cos(2x) gives us

cos(2x) = 1 not 0, so you can start by finding theta where cos(theta) = 1

Answer 34

[1-cos(2x)] = 0
cos(2x) = 1


um 0

Answer 35

good, only one x value this time

dividing that by 2 gives us 0 again

0 + 2pi*n

which is already covered w/ sin

so it really comes down to

0 + 2pi*n and pi/2 + 2*pi*n

Answer 36

ok there's something that's still not quite sitting right with me on this one

I believe there also needs to be pi + 2pi*n to account for odd pi multiples