What is the general solution to $\sin\theta=\frac12$?
I have an incorrect solution but I don't know why. \begin{align*} \sin\theta & =\frac12\\ \sin\theta & =\sin\alpha\\ \alpha & =\arcsin\left(\frac12\right)=\frac{\pi}{6}\\ \theta & =n\pi+\alpha(-1)^n\\ \theta & =n\pi+\frac{\pi(-1)^n}6\\ \end{align*} I then found the two general solutions for when $n$ is even and odd.
$$\theta=\pi(n+\frac16)$$ and $$\theta=\pi(n-\frac16)$$
What am I doing wrong here?
Thanks
I don't think there is anything particularly wrong with your answer, however it is more common to see it written without conditions on $n$ like you have put.
Your answer was:
$$\{\pi(n+\frac{1}{6})~:~n~\text{is an even integer}\}\cup \{\pi(n-\frac{1}{6})~:~n~\text{is an odd integer}\}$$
I would personally have written it in the following way:
$$\{2\pi n +\frac{\pi}{6}~:~n~\text{is any integer}\}\cup \{2\pi n + \frac{5\pi}{6}~:~n~\text{is any integer}\}$$
To see that they are equivalent, note that for $n$ even, you can rewrite it as $n=2k$. You have then for all even $n$, $\pi(n+\frac{1}{6}) = \pi (2k + \frac{1}{6}) = 2\pi k + \frac{\pi}{6}$. By relabeling you see that the set on the left in both your and my answers are the same.
(notice further that to cover all even integers $n$ is equivalent to covering all integers $k$ for $n=2k$., i.e. there is a bijection between them)
For the set on the right, noting that for $n$ odd, it can be written as $n=2k+1$, you have then $\pi(n-\frac{1}{6}) = \pi(2k+1 - \frac{1}{6}) = \pi(2k+\frac{5}{6}) = 2\pi k + \frac{5\pi}{6}$. Again, by relabeling $k$ to $n$ you see our answers agree once again. Notice further that both of our answers agree with the answers given in the other posts and with wolfram.alpha
Looking at the trigonometric circle, angles of a given sine are found on an horizontal line. They are supplementary and indeterminate by a number of full turns.
Assuming you know a particular solution, $\theta_0$, all solutions are of the form $$2k\pi+\theta_0$$ or $$2k\pi+(\pi-\theta_0)=(2k+1)\pi-\theta_0,$$which is the same result as yours. If you like, you can indeed summarize as $$n\pi+(-1)^n\theta_0.$$
$$\sin \alpha = \sin x$$ $$\sin \alpha = \frac12$$ $$\alpha = 2k\pi + x $$ $$\alpha = 2k\pi + \pi -x$$ $$x = \frac\pi6$$ Now you can plug $x$ in the formulas above and find out the general solutions for your problem.
You only use one symmetry in your first attempt.
The first symmetry is $\sin(x) = \sin(x + 2\pi)$: $$ x = 2\pi k + \arcsin(1/2) = 2 \pi k + \pi/6 = \pi(2k + 1/6) $$ The second symmetry is $\sin(x) = - \sin(x + \pi)$: $$ x = 2\pi k + \pi - \arcsin(1/2) = 2 \pi k + \pi - \pi/6 = \pi(2k + 5/6) $$ for $k \in \mathbb{Z}$.
