Question

Help solving this identity? Medal and fan

2sin(2x)-2sinx +2sqrt3cosx-sqrt3=0

Thanks(:

Answer 1

sin(2x) = ?

Answer 2

equation
Well, it may be an identity, but I don't think so.

Answer 3

If it were an identity it would be true for all values of x

Answer 4

Just wrongly phrased.

Answer 5

the fact that left side expression can be easily factored, makes this quite easy to solve.

ofcourse after applying the formula for sin 2x :)

Answer 6

So the first thing I would do is turn sin(2x) into 2 sin(x)cos(x)

2 (2 sin(x) cos(x) ) - 2 sin(x) + 2 Sqrt[3] cos(x) - Sqrt[3] = 0.
4 sin(x) cos(x) - 2 sin(x) + 2 Sqrt[3] cos(x) - Sqrt[3] = 0.

Answer 7

Mira, let the user do it herself! I don't find any reason to put all of our work unless the user makes an effort.

Answer 8

Sorry, I was afk. Yeah, I didn't mean to put identity, It's 3:24 am, a bit tired.

Right, I have already done that in my work.

Answer 9

Okay, so, is that what you have done yet? or something else too?

Answer 10

Yep, I have that.

Answer 11

I then took a 2sinx out...would that be right to do?

Answer 12

Oh wait, nvm, I shouldn't do that.

Answer 13

What direction should I be going with this? I need a push forward, I already had all the information that people had given.

Answer 14

Let us start from the beginning.

What is the formula for sin(2x) = ?

Answer 15

sin(2x) = 2sinxcosx

Answer 16

Good. Now, put this in the equation we have.

Answer 17

(4sinxcosx-2sinx+ 2sqrt3cosx-sqrt3 = 0

Answer 18

Good. now,

let us try to factor it out..

the last two terms have something common in them, what is it?

Answer 19

the sqrt3.

Answer 20

Good work! Now, in the first two terms, what is common?

Answer 21

the 2sinx

Answer 22

Good work again...!

Now, try to do the factoring. Can you do it?

Answer 23

(sqrt3)(2sinx)(2sinxcosx+2cosx-2) so far correct?

Answer 24

No! This is not the correct way to do the factoring.

See, we have 2sin x common int he first 2 terms and sqrt(3) common in the last 2 terms :

(4sinxcosx-2sinx+ 2sqrt3cosx-sqrt3 = 0

\(2\sin x ( 2\cos x - 1) + \sqrt{3} (2\cos x - 1) =0 \)

Are you getting it?

Answer 25

Yeah, I get that sorry. Factoring with trig functions always throws me off.

Answer 26

No worries!

So, just take 2cos(x) - 1 common now.

Answer 27

Hold on, how did you get rid of the other 2sinx attached to the cos when it was 4sinx?

Answer 28

\(4\sin x \cos x - \sin x + 2\sqrt{3} \cos x - \sqrt{3} = 0\)

See,

the first two terms have 2 sin x common :

So,

if you take 2 sin x common from 4 sin x cos x

you get : 2 cos x

Answer 29

Thus, I wrote it as:

2sinx (2cos -1) + sqrt(3) (2 cos x-1)= 0

Answer 30

That still doesn't account for the 4 though..taking 2sinx out leaves 2sinxcosx...unless you converted the 2sinx into 2cosx-1, but then you also have the cosx still left there to multiply?

Answer 31

(2sinxcosx) = 2sinx ((2cosx-1)cosx-1) ?

Answer 32

Let us take only 2 terms now :

4 sin x cos x - sinx

Now,

I can write 4sinx cosx as : \(2\sin x \times 2\cos x \)

Right?

Answer 33

OH. I was only connecting that with the sin only. That makes much more sense.

Answer 34

Oops wait, I wrote that wrng

it is 4sinx cos x - 2sinx

Answer 35

Now, we have :

\(\color{blue}{2 \sin x} \times 2\cos x - \color{blue}{2\sin x}\)

Take 2sinx common now..

Answer 36

Right.

Answer 37

So, what do you get?

Answer 38

Well, I'm guessing we don't pay any mind to the sqrt 3, so I have 2sinx and cosx= 1/2

Answer 39

No...! Don't jump directly to the conclusion, as this may lead to big mistakes.

We have :

\(2\sin x ( 2\cos x - 1) + \sqrt{3} (2\cos x -1 )=0\)

Take \(2\cos x - 1\) common.

Answer 40

So the first thing I would do is turn sin(2x) into 2 sin(x)cos(x)

2 (2 sin(x) cos(x) ) - 2 sin(x) + 2 Sqrt[3] cos(x) - Sqrt[3] = 0.
4 sin(x) cos(x) - 2 sin(x) + 2 Sqrt[3] cos(x) - Sqrt[3] = 0.

then I would factor by grouping.
2 sin(x) (2 cos(x) - 1) + Sqrt[3] (2 cos(x) - 1) = 0
Therefore
(2 sin(x) + sqrt[3])(2 cos(x) - 1) = 0.
This then is easily solvable for the roots.
cos(x) = 1/2
So this means +- Pi/3 + 2 * n * Pi where n is any integer.
sin(x) = - Sqrt[3]/2. This means 4pi/3 + 2 * n * Pi, n any integer. We ignore the other solution 5pi/3 + 2 * n * Pi because it's redundant with what was found from the other root.

Answer 41

Make sense? :) @seehearcreate

Answer 42

Oh, you multiply the sinx by sqrt 3..... Yeah, that makes sense(:

Answer 43

yup! =)

Answer 44

Sorry, but I don't appreciate someone jumping in and writing the whole direct solution.

Anyways, no worries, I will leave. Good luck @seehearcreate !

Answer 45

@mathslover You helped me the most with this problem, so I'll medal you, but I'll fan you both(: Thankyou, both of you!

Answer 46

You're welcome! =)