# How would I solve sin2Ѳ+sin3Ѳ=0?

1. Sep 1, 2014

### beingandfluffy

EDIT: Please delete...was very confused and have now found solution!

Last edited: Sep 1, 2014
2. Sep 2, 2014

3. Sep 2, 2014

### beingandfluffy

Sure!
Use sin(A+B)/2*cos(A-B)/2
=2sin (3x+2x)/2*cos (2x-3x)/2
2sin(5x)=0
Thus sinx=0
x=0, 180+2 (pi)n
2cos (-x)/2
Thus cosx=1/2
x=90+2 (pi)n

4. Sep 2, 2014

### ehild

Do you mean that Θ=pi/2 is a solution?
But sin(2Θ)+sin(3Θ)=sin(180°)+sin(270°)=-1.

And what about Θ=72°? Is not it a solution, too?

ehild

5. Sep 2, 2014

### beingandfluffy

Well, the solutions manual is getting:
Θ=0, 2∏/5, 4∏/5, ∏, 6∏/5, 8∏/5
The interval is also 0 ≤ θ ≤2∏

I forgot to include the interval before, but could figure it out without the interval. Now I don't know where to go with the problem as it looks like I'm solving it completely wrong.

Last edited: Sep 2, 2014
6. Sep 2, 2014

### ehild

No, you started very well, it was a splendid idea.
But you miss some parentheses. It should be 2sin [(3x+2x)/2]*cos [(2x-3x)/2]
So sin(5x/2)=0, and from this does not follow that sinx=0. Instead, sin(5x/2)=0 that is, 5x/2=kπ. x=???

Do the same with the other factor, cos(x/2).

ehild