# Trig equation (and its solutions)

• Sep 25th 2008, 08:38 AM
courteous
Trig equation (and its solutions)
(Should be) simple trigonometric equation: $\displaystyle \sin3x=\sin6x$.

Given solutions [1] $\displaystyle x=\frac{k\pi}{3}$, [2a] $\displaystyle x=\frac{\pi}{9}+\frac{2k\pi}{3}$, [2b] $\displaystyle x=-\frac{\pi}{9}+\frac{2k\pi}{3}$

Why is it, that with 'sum (difference) theorem' $\displaystyle \sin6x-\sin3x=0 \rightarrow 2\cos\frac{9x}{2}\sin\frac{3x}{2}=0$, I get different solutions than with $\displaystyle sin(3x) - sin(6x) = sin(3x) - 2sin(3x)cos(3x) = sin(3x)[1-2cos(3x)] = 0$ (at least it seems so) (Headbang)
• Sep 25th 2008, 08:54 AM
Showcase_22
This is what I did:

$\displaystyle sin3x=sin6x$

$\displaystyle 3x=6x+2\pi n$ or $\displaystyle 3x=(2n+1)\pi-6x$

Now we have two equations to solve:

$\displaystyle 3x=-2\pi n$

$\displaystyle 9x=(2n+1)\pi$

$\displaystyle x=\frac{\pi}{9}(2n+1)$ and $\displaystyle x=\frac{-2\pi n}{3}$

I would do it this way except I don't get the solutions you noted down. I think I did it right...... :s