Solving for sinΘ in sinΘ=sin2Θ

[beingandfluffy]

Homework Statement

sinΘ=sin2Θ

Homework Equations

Double-angle formula...
sin2Θ=2sinΘcosΘ

sinΘ=2sinΘcosΘ
sinΘ

The Attempt at a Solution

sinΘ=2sinΘcosΘ
sinΘ-2sinΘcosΘ=0
sinΘ(1-2cosΘ)=0

Now I'm getting stuck because 1-2cosΘ is not an identity of any sort.

Any help would be appreciated...thanks!

 

[Nathanael]

sinΘ=2sinΘcosΘ

Instead of subtracting; what about dividing both sides by sinΘ?


Edit:

sinΘ(1-2cosΘ)=0

You could divide right here, too. (zero divided by anything* is still zero)


*anything except zero

 

[beingandfluffy]

I'll try that!
Edit:
AH I see it

 

[Mark44]

Instead of subtracting; what about dividing both sides by sinΘ?

This is not a good idea. Dividing by sinθ causes you to lose solutions for which sinθ = 0.

Edit:

You could divide right here, too. (zero divided by anything* is still zero)


*anything except zero

 

[Nathanael]

Actually, your way is better

sinΘ(1-2cosΘ)=0

It is better to set it up like this because this shows all of the solutions. (I overlooked this in my first post.)When is sinΘ(1-2cosΘ) equal to zero? (There are two solutions)
edit:

This is not a good idea. Dividing by sinθ causes you to lose solutions for which sinθ = 0.

Yes I just realized this thank you

edit#2:

There are LOTS more than two solutions...

I had a feeling you would point this out too I couldn't think of a concise way to say it. (I meant two ways)
Maybe I should stay out of the math section

 

[Mark44]

Homework Statement

sinΘ=sin2Θ

Homework Equations

Double-angle formula...
sin2Θ=2sinΘcosΘ

sinΘ=2sinΘcosΘ
sinΘ

The Attempt at a Solution

sinΘ=2sinΘcosΘ
sinΘ-2sinΘcosΘ=0
sinΘ(1-2cosΘ)=0

Now I'm getting stuck because 1-2cosΘ is not an identity of any sort.

Doesn't need to be. You're on the right track. If the product above equals zero, that means that either sinθ = 0 or 1 - 2cosθ = 0.

Any help would be appreciated...thanks!

 

[Mark44]

Actually, your way is better

It is better to set it up like this because this shows all of the solutions. (I overlooked this in my first post.)


When is sinΘ(1-2cosΘ) equal to zero? (There are two solutions)

There are LOTS more than two solutions...

edit:

Yes I just realized this thank you

 

[nrqed]

Homework Statement

sinΘ=sin2Θ

Homework Equations

Double-angle formula...
sin2Θ=2sinΘcosΘ

sinΘ=2sinΘcosΘ
sinΘ

The Attempt at a Solution

sinΘ=2sinΘcosΘ
sinΘ-2sinΘcosΘ=0
sinΘ(1-2cosΘ)=0

Now I'm getting stuck because 1-2cosΘ is not an identity of any sort.

Any help would be appreciated...thanks!

You have the product of two functions of Θ. The result is zero when either sinΘ or 1-2cosΘ is zero.
This occurs at what angles? (As for 1-2cosΘ=0, just isolate the angle. Do you see how to do that?)

 

[beingandfluffy]

Let's see...
I could do sinΘ=0 or 1-2cosΘ=0
sinΘ=0
Θ= (pi)/2
1-2cosΘ=0
2cosΘ=-1
cosΘ=-1/-2
cosΘ=1/2
Θ=5(pi)/3 or (pi)/3
I see that if I do it this way, instead of dividing out the sine, I capture one more solution.
Very helpful...thank you!

 

[Mark44]

Since there are no restrictions on θ (or at least none shown in your OP), you should add integer multiples of 2π to each of your solutions to capture all possible solutions.

 

[nrqed]

Let's see...
I could do sinΘ=0 or 1-2cosΘ=0
sinΘ=0
Θ= (pi)/2
1-2cosΘ=0
2cosΘ=-1
cosΘ=-1/-2
cosΘ=1/2
Θ=5(pi)/3 or (pi)/3
I see that if I do it this way, instead of dividing out the sine, I capture one more solution.
Very helpful...thank you!

Watch out, $\sin(\theta) =0$ at $\theta = \ldots, - \pi, 0 , \pi, \ldots$,

not at $\theta = \pi/2$!

 

[BvU]

Watch out some more !

Let's see...
I could do sinΘ=0 or 1-2cosΘ=0
sinΘ=0
Θ= (pi)/2
1-2cosΘ=0
2cosΘ=-1
cosΘ=-1/-2
cosΘ=1/2
Θ=5(pi)/3 or (pi)/3
I see that if I do it this way, instead of dividing out the sine, I capture one more solution.

sinΘ(1-2cosΘ)=0 $\Leftrightarrow$ sinΘ=0 $\vee$ (1-2cosΘ)=0

First one: sinΘ=0 $\Leftrightarrow$ sinΘ=sin(0) $\Leftrightarrow$ Θ = 0 + 2n∏ $\vee$ Θ = ∏ - 0 + 2n∏ $\Leftrightarrow$ Θ = n∏ (with n an integer)

Now you do the second one. And don't write 1-2cosΘ=0 on one line and 2cosΘ=-1 on the next !

Being meticulous while learning this pays off all through the rest of your life (and that of possible pupils/students you might have later on )

By the way, welcome to PF !