Question

3sinx=3cos2x

Answer 1

3sinx=3cos2x
sinx=cos2x
sinx=cos(x+x)
sinx=cos(x)cos(x)-cos(x)cos(x)
sinx=cos²x-cos²x
sinx=0

Never take people seriously, unless yu are sure about them.

Answer 2

okay but what about cosine? and yeah I know some people are just mean.

Answer 3

FOR REAL!

3sinx=3cos2x
sinx = cos(2x)
sinx - cos(2x)=0

RULE: cos 2x = 1 - 2 (sin x) ^ 2

sin(x)-[ 1 - 2 (sin x) ^ 2 ] = 0
2 (sin x) ^ 2 + sin(x) -1 = 0

take it from here...

Answer 4

okay this is what I did but I am not sure if its correct.
sin4x=cos2x
2sin(2x)cos(2x)=cos(2x)
(2 sin(2x)-1)cos(2x)=0
sin(2x)=1/2 cos(2x)=0

Answer 5

sin(4x) ?

Answer 6

Are you doing a new problem, or the initial one ?

Answer 7

wow I didn't even notice! I typed it wrong to start with. sorry :(

Answer 8

it's okay.

Answer 9

so is that fine?

Answer 10

yes, yes. You can look at what did for this problem. Or perhaps you have a different approach. Whetaver you prefer.

Answer 11

okay thank you so much!

Answer 12

You already got the answer? Can you share please? (Making sure...)

Answer 13

yeah aren't the answers 7pi/4 pi/4 pi/12 etc.

Answer 14

Using the quadratic formula, I am getting

sin x = -1 or sin x = 1/2

Answer 15

your interval is (-∞ , ∞)
Like an example of an answer would be
π/6.

Answer 16

im confused now I thought I was doing it right

Answer 17

Can you post(or describe in couple steps) your work please?

Answer 18

its already up there ^^^

Answer 19

Not really, you started wit the wrong thing (sin 4x), if you are referring to your last post where you say

────────────────────────────
okay this is what I did but I am not sure if its correct.
sin4x=cos2x
2sin(2x)cos(2x)=cos(2x)
(2 sin(2x)-1)cos(2x)=0
sin(2x)=1/2 cos(2x)=0
────────────────────────────

Answer 20

yeah that's were I got my answers from

Answer 21

But that's not the right problem, your problem is

sin(x)=cos(2x)

Answer 22

no my problem is 3sin 4x=3cos2x that's why I said that I typed it wrong because I forgot the 4

Answer 23

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3sinx=3cos2x
sinx = cos(2x)
sinx - cos(2x)=0

USE THE RULE: cos (2x) = 1 - 2 (sin x) ^ 2

sin(x) - [ 1 - 2 (sin x) ^ 2 ] = 0 re-writing in descending order...

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

2a^2+a-1=(2a-1)(a+1)
a=1/2 , -1

hence, sin(x)=1/2 and sin (x)=-1


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Answer 24

I have to go and study now but thank you for your help!

Answer 25

Don;t think that I helped, but yw -:(