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In a previous question I asked about how to apply $\operatorname{arctan2}$ to:

If $\sin(\theta) = \frac{-1}{2}$ and $\cos(\theta) = \frac{\sqrt{3}}{2}$ which is found as: $\operatorname{arctan2}( \frac{\sqrt{3}}{2} , \frac{-1}{2}) \implies \theta = \frac{-\pi}{6}$ or $-30^\circ$

Now I am asked to find:

If $\sin(\phi)\sin(\theta) = 0.2$ and $\sin(\phi) \cos(\theta) = -0.3$ and $\sin(\phi) > 0$ then, what is $\theta$? Repeat for $\sin(\phi) < 0$

I am confused as what I should do with the $\sin(\phi)$.

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    @J. M.: that is why I had the $\theta+2k\pi$. I guess I could have said $\theta\pmod{2\pi}$, but that might be interpreted as being in $[0,2\pi)$.2011-09-24

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Hint: You can divide out the $\sin \phi$ between the two equations, leaving just an equation for $\tan \theta$.

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    @J.M.: Yes, the title did talk about arctan2, so the $\sin \phi$ can confuse the signs.2011-09-25