solving trigonometric equations

**nathanand** · March 15th 2009, 02:22 PM

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

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

**e^(i*pi)** · March 15th 2009, 02:41 PM

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 $2A = \frac{2\pi}{3} <br /> \rightarrow A = \frac{\pi}{3}$

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

**nathanand** · March 15th 2009, 03:03 PM

Thanks for the help!

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