# Solutions between (0,2pi]

• March 21st 2011, 09:03 AM
PintsofGuinness
Solutions between (0,2pi]
I am just trying to get some help on how to get this equation right. I have it equal to 0. Anything will help thanks.

sin5x-sinx=2cos3x
sin4x-2cos3x=0 (I think this is correct, just need help moving on to the next step.
• March 21st 2011, 10:06 AM
FernandoRevilla
Quote:

Originally Posted by PintsofGuinness
sin5x-sinx=2cos3x
sin4x-2cos3x=0

What?.

$\sin A-\sin B=2\cos\dfrac{A+B}{2}\sin \dfrac{A-B}{2}$

So,

$\sin 5x-\sin x=2\cos 3x \sin 2x$