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

[Speedking96]

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.

 

[HallsofIvy]

Are you to find x in the interval $0\le x< 2\pi$? It would be better if you would tell us that!
Yes, sin(x)= 0 for x= 0 and $x= \pi$.

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 $2x= \pi/6, \pi= \pi/6= 5\pi/6, 2\pi+ \pi/6= 13\pi/6, 3\pi- \pi/6= 17\pi/6$

Notice that I have gone to 0 to $2\pi$ because I am going to divide by 2:

$x= \pi/12$, $x= 5\pi/12$, $13\pi/12$, $17\pi/12$.

 

[Speedking96]

Are you to find x in the interval $0\le x< 2\pi$? It would be better if you would tell us that!
Yes, sin(x)= 0 for x= 0 and $x= \pi$.

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 $2x= \pi/6, \pi= \pi/6= 5\pi/6, 2\pi+ \pi/6= 13\pi/6, 3\pi- \pi/6= 17\pi/6$

Notice that I have gone to 0 to $2\pi$ because I am going to divide by 2:

$x= \pi/12$, $x= 5\pi/12$, $13\pi/12$, $17\pi/12$.

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.

 

[Ray Vickson]

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.

 

[Speedking96]

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

 

[Speedking96]

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?

 

[Ray Vickson]

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.

 

[SammyS]

...

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

Notice that I have gone to 0 to $2\pi$ because I am going to divide by 2:

...

I believe Halls meant to write

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

in particular, $\ ... \pi- \pi/6= 5\pi/6,\ ...$

 

[Speedking96]

I believe Halls meant to write

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

in particular, $\ ... \pi- \pi/6= 5\pi/6,\ ...$

Yes. I have understood the problem.