I have a doubt with regards to the general solution of $\sin(\theta) = -1/2$.
I know that the principal solutions are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.
So my main question is: Should the general solution be $n\pi$ + $(-1)^{n}$$\frac{7\pi}{6}$ or $n\pi$ + $(-1)^{n}$$\frac{11\pi}{6}$?
Apparently, WolframAlpha uses $\frac{7\pi}{6}$ for the general solution. My apologies to you if you found this question a bit silly. I appreciate your time and effort.
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$\begingroup$There is no difference, since$$3\pi+(-1)^3\frac{7\pi}6=\frac{11\pi}6\quad\text{and}\quad3\pi+(-1)^3\frac{11\pi}6=\frac{7\pi}6.$$So, you get the same solutions in both cases.
$\endgroup$ 3 $\begingroup$See if this helps:
We know $\sin{\theta}$ is negative in quadrants 3 & 4:
Therefore, for Quadrants 3 & 4, these formulas can be used respectively:$$\text{Quadrant 3: } (2n + 1)\pi + \theta$$$$\text{Quadrant 4: } 2\pi(n + 1) - \theta$$where $n$ denotes, the rotation (repeat; as the cycle repeats ever $2\pi$) count and $\theta$ is, well, your angle (which, the base is, $\frac{\pi}{6}$)
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