Finding Solutions to sin(2x)=0 and 2sin(2x)-1=0

  • Thread starter Speedking96
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In summary, this conversation is discussing how to solve the equation (sin(2x)) * (2sin(2x)-1)=0. The first solution is x= 0 or x= π, and the second solution is x= π/12 ; 11π/12. There may be two more solutions, but it is not clear from the given information. The conversation also touches on the importance of clarifying the interval in which to find x and the proper way to remove the "2" from sin(2x) when solving the equation.
  • #1
Speedking96
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0

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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  • #2
Are you to find x in the interval [itex]0\le x< 2\pi[/itex]? It would be better if you would tell us that!
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 cannot then declare that sin(x)= 1/4!
The "2" is inside the function- you cannot divide by 2 until after you have removed the sine.
(That's a howler of an error! I really hope that was carelessness.)

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
HallsofIvy said:
Are you to find x in the interval [itex]0\le x< 2\pi[/itex]? It would be better if you would tell us that!
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 cannot then declare that sin(x)= 1/4!
The "2" is inside the function- you cannot divide by 2 until after you have removed the sine.
(That's a howler of an error! I really hope that was carelessness.)

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].

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
Speedking96 said:
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
Ray Vickson said:
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.

Yes, I meant to write 2x = pi/6 not sin 2x = pi/6
 
  • #6
I mainly don't understand the part where HallsOfIvy wrote pi = pi/6 = 5pi / 6 ...

Where did he get the pi = pi/6 = 5pi / 6 ... from?
 
  • #7
Speedking96 said:
I mainly don't understand the part where HallsOfIvy wrote pi = pi/6 = 5pi / 6 ...

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
HallsofIvy said:
...

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
SammyS said:
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.
 

What is the first step in solving the equation sin(2x)=0?

The first step is to isolate the sine function by dividing both sides of the equation by 2, resulting in sin(x)=0.

How many solutions are there for the equation sin(2x)=0?

There are an infinite number of solutions for this equation, as the sine function has a period of 2π and repeats itself infinitely. Therefore, there will be multiple values of x that satisfy the equation.

What is the general solution for sin(2x)=0?

The general solution for this equation is x = nπ, where n is any integer. This means that any value of x that is a multiple of π will satisfy the equation.

What is the first step in solving the equation 2sin(2x)-1=0?

The first step is to isolate the sine function by subtracting 1 from both sides of the equation, resulting in 2sin(2x)=1.

How many solutions are there for the equation 2sin(2x)-1=0?

There are two solutions for this equation, as the sine function oscillates between -1 and 1. Therefore, there will be two values of x that satisfy the equation.

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