# Math Help - solving trigonometric equations

1. ## solving trigonometric equations

I've been stuck trying to figure this one out for a while...

sin(2A) + sin(4A) = 0

2. Originally Posted by nathanand
I've been stuck trying to figure this one out for a while...

sin(2A) + sin(4A) = 0
Let $u = 2A$

$sin(u) + sin(2u) = 0$

$2sin(u)cos(u) + sin(u) = 0$

$sin(u)(2cos(u)+1) = 0$

Either sin(u) = sin(2A) = 0 so A=0

or $2cos(u) = -1 \rightarrow cos(u) = -\frac{1}{2}$

and so $u = \frac{2\pi}{3}$ thus $2A = \frac{2\pi}{3}
\rightarrow A = \frac{\pi}{3}$

bear in mind this graph will repeat every $\pi$ radians instead of every $2\pi$ radians

3. Thanks for the help!