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## Homework Statement

(sin(2x)) * (2sin(2x)-1)=0

**2. The attempt at a solution**

sin (2x) = 0

sinx=0

x= 0, π

2sin(2x)-1=0

sin (2x) = 1/2

sinx = 1/4

x = π/12 ; 11π/12

There are two more solutions but I cannot seem to find them.

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- Thread starter Speedking96
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- #1

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(sin(2x)) * (2sin(2x)-1)=0

sin (2x) = 0

sinx=0

x= 0, π

2sin(2x)-1=0

sin (2x) = 1/2

sinx = 1/4

x = π/12 ; 11π/12

There are two more solutions but I cannot seem to find them.

- #2

HallsofIvy

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Yes, sin(x)= 0 for x= 0 and [itex]x= \pi[/itex].

For 2 sin(2x)- 1= 0, yes, that leads to sin(2x)= 1/2.

But you

The "2" is

(That's a

From sin(2x)= 1/2 you get [itex]2x= \pi/6, \pi= \pi/6= 5\pi/6, 2\pi+ \pi/6= 13\pi/6, 3\pi- \pi/6= 17\pi/6[/itex]

Notice that I have gone to 0 to [itex]2\pi[/itex] because I am going to divide by 2:

[itex]x= \pi/12[/itex], [itex]x= 5\pi/12[/itex], [itex]13\pi/12[/itex], [itex]17\pi/12[/itex].

- #3

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Yes, sin(x)= 0 for x= 0 and [itex]x= \pi[/itex].

For 2 sin(2x)- 1= 0, yes, that leads to sin(2x)= 1/2.

But youcannotthen declare that sin(x)= 1/4!!

The "2" isinsidethe function- you cannot divide by 2 until after you have removed the sine.

(That's ahowlerof an error! I really hope that was carelessness.)

Fromsin(2x)= 1/2 you get [itex]2x= \pi/6, \pi= \pi/6= 5\pi/6, 2\pi+ \pi/6= 13\pi/6, 3\pi- \pi/6= 17\pi/6[/itex]

Notice that I have gone to 0 to [itex]2\pi[/itex] because I am going to divide by 2:

[itex]x= \pi/12[/itex], [itex]x= 5\pi/12[/itex], [itex]13\pi/12[/itex], [itex]17\pi/12[/itex].

My teacher had told me that I could remove the two from sin(2x) right away... sorry about that.

From the sin(2x) = pi/6 ... i am kind of lost

I know how you got pi/6, but the rest, I don't quite understand.

- #4

Ray Vickson

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My teacher had told me that I could remove the two from sin(2x) right away... sorry about that.

From the sin(2x) = pi/6 ... i am kind of lost

I know how you got pi/6, but the rest, I don't quite understand.

I hope that what your teacher means is that if you have sin(2x) = 0, then setting y = 2x you have sin(y) = 0, NOT sin(x) = 0. In fact, if sin(2x) = 0 then sin(x) is as far from 0 as the 'sin' can get: we would have either sin(x) = +1 or sin(x) = -1.

Also, what you wrote above makes no sense: you do not have sin(2x) = pi/6; you do have sin(2x) = 1/2 = sin(pi/6), so 2x = pi/6.

Last edited:

- #5

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I hope that what your teacher means is that if you have sin(2x) = 0, then setting y = 2x you have sin(y) = 0, NOT sin(x) = 0. In fact, if sin(2x) = 0 then sin(x) is as far from 0 as the 'sin' can get: we would have either sin(x) = +1 or sin(x) = -1.

Also, what you wrote above makes no sense: you do not have sin(2x) = pi/6; you do have sin(2x) = 1/2 = sin(pi/6), so2x = pi/6.

Yes, I meant to write 2x = pi/6 not sin 2x = pi/6

- #6

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Where did he get the pi = pi/6 = 5pi / 6 ... from?

- #7

Ray Vickson

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Where did he get the pi = pi/6 = 5pi / 6 ... from?

That is NOT what he wrote. Read it again.

He is just listing the several possible values of x, so what he wrote is shorthand for x = 0 or x = pi/6 or x = 5pi/6 or .... . He is not saying those values are equal to each other.

- #8

SammyS

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...

From sin(2x)= 1/2 you get [itex]2x= \pi/6, \pi= \pi/6= 5\pi/6, 2\pi+ \pi/6= 13\pi/6, 3\pi- \pi/6= 17\pi/6[/itex]

Notice that I have gone to 0 to [itex]2\pi[/itex] because I am going to divide by 2:

...

I believe Halls meant to write

From sin(2x)= 1/2 you get [itex]2x= \pi/6,\ \pi- \pi/6= 5\pi/6,\ 2\pi+ \pi/6= 13\pi/6,\ 3\pi- \pi/6= 17\pi/6\ .[/itex]

in particular, [itex]\ ... \pi- \pi/6= 5\pi/6,\ ...[/itex]

- #9

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I believe Halls meant to write

From sin(2x)= 1/2 you get [itex]2x= \pi/6,\ \pi- \pi/6= 5\pi/6,\ 2\pi+ \pi/6= 13\pi/6,\ 3\pi- \pi/6= 17\pi/6\ .[/itex]

in particular, [itex]\ ... \pi- \pi/6= 5\pi/6,\ ...[/itex]

Yes. I have understood the problem.

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