<s>[INST] How to Solve for $x$ in the equation $\sin(2x) = -\frac{1}{2}$ within the interval of $[0, 2\pi]$? So far, I only have the solutions $11\pi/12$, and $7\pi/12$ (I think there are a total of $4$). I think this has something to do with trig identities, and should be able to be solved without using a calculator. However, I have only been able to get the above solutions using my calculator.
 [/INST] If $$\sin(2x) = -\frac{1}{2}$$ then 
\begin{align*}
2x & = \arcsin\left(-\frac{1}{2}\right) + 2n\pi & 2x & = \pi - \arcsin\left(-\frac{1}{2}\right) + 2m\pi\\
2x & = -\frac{\pi}{6} + 2n\pi & 2x & = \pi - \left(-\frac{\pi}{6}\right) + 2m\pi\\
x & = -\frac{\pi}{12} + n\pi & 2x & = \frac{7\pi}{6} + 2m\pi\\
 & & x & = \frac{7\pi}{12} + m\pi
\end{align*} 
where $m, n \in \mathbb{Z}$.  The requirement that $x \in [0, 2\pi]$ implies $n = 1, 2$ and $m = 0, 1$.  
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