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

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

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- Mar 15th 2009, 02:22 PMnathanandsolving trigonometric equations
I've been stuck trying to figure this one out for a while...

sin(2A) + sin(4A) = 0 - Mar 15th 2009, 02:41 PMe^(i*pi)
Let $\displaystyle u = 2A$

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

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

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

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

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

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

\rightarrow A = \frac{\pi}{3}$

bear in mind this graph will repeat every $\displaystyle \pi$ radians instead of every $\displaystyle 2\pi$ radians - Mar 15th 2009, 03:03 PMnathanand
Thanks for the help!