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I'm trying to set a simple trig equation equal to zero, for use with the first derivative test, etc. I have:

$-2(\sin\theta+\sin2\theta)$

So, I need to get

$\sin\theta=-\sin2\theta$

I'm drawing a complete blank here.

If it matters, but it shouldn't the original equation I've derived from is:

$f(\theta)=2\cos\theta+\cos2\theta$

I guess what I'm really asking for is a refresher in solving trig equations, if someone doesn't mind. :)

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Note that $\sin(2\theta)=2\sin(\theta)\cos(\theta)$.

Wikipedia's list of trig identities should prove very helpful :)

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    $\begingroup$ Great, thanks! I gave up too early on the simplification! $\endgroup$ – Josh Oct 14 '11 at 2:55
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    $\begingroup$ And remember: you can't divide both sides of the equation by $\sin\theta$ unless $\sin\theta \ne 0$. So you need to say: Either $\sin\theta=0$ or $\dots\dots$. $\endgroup$ – Michael Hardy Oct 14 '11 at 5:24

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