# Math Help - Triple angle trig function

1. ## Triple angle trig function

Solve:
sin3xcosx = cos3xsinx

Here's what I've done so far:

sin3xcosx = cos3xsinx
sin3xcosx - cos3xsinx = 0
sin(2x+x)cosx - cos(2x+x)sinx = 0
(sin2xcosx + cos2xsinx)cosx - (cos2xcosx - sin2xsinx)sinx = 0

Here's where I'm stuck. Should I distribute the cosx and sinx on either side? Or plug in the sin2x/cos2x double angle identities?
Any help is appreciated!

2. Originally Posted by live_laugh_luv27
Solve:
sin3xcosx = cos3xsinx
I would write this as $\sin3x\cos x - \cos3x\sin x = 0$ and use the formula $\sin A\cos B - \cos A\sin B = \sin(A-B)$.

3. So, what should I do after I simplify the equation to sin(3x - x) = 0?

4. Originally Posted by live_laugh_luv27
So, what should I do after I simplify the equation to sin(3x - x) = 0?
3x – x = 2x, so your equation is sin(2x) = 0.