# How would I solve sinѲ=sin2Ѳ?

1. Sep 1, 2014

### beingandfluffy

1. The problem statement, all variables and given/known data

sinΘ=sin2Θ

2. Relevant equations

Double-angle forumla...
sin2Θ=2sinΘcosΘ

sinΘ=2sinΘcosΘ
sinΘ

3. 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!

2. Sep 1, 2014

### Nathanael

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

*anything except zero

3. Sep 1, 2014

### beingandfluffy

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

4. Sep 1, 2014

### Staff: Mentor

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

5. Sep 1, 2014

### Nathanael

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:
Yes I just realized this thank you

edit#2:
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 :tongue:

Last edited: Sep 1, 2014
6. Sep 1, 2014

### Staff: Mentor

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.

7. Sep 1, 2014

### Staff: Mentor

There are LOTS more than two solutions...

8. Sep 1, 2014

### nrqed

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?)

9. Sep 1, 2014

### 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.

10. Sep 1, 2014

### Staff: Mentor

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.

11. Sep 1, 2014

### nrqed

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

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

12. Sep 2, 2014

### BvU

Watch out some more !
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 !