EDIT: Please delete...was very confused and have now found solution!
How about posting the answer instead, then someone else confused can benefit from your confusion ;)
=2sin (3x+2x)/2*cos (2x-3x)/2
x=0, 180+2 (pi)n
Do you mean that Θ=pi/2 is a solution?
And what about Θ=72°? Is not it a solution, too?
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.
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).
Separate names with a comma.