What is the general solution to $\sin\theta=\frac12$? 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
 A: 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
A: 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 $$n\pi+(-1)^n\theta_0.$$
A: $$\sin \alpha = \sin x$$
$$\sin \alpha = \frac12$$
$$\alpha = 2k\pi + x $$
$$\alpha = 2k\pi + \pi -x$$
$$x = \frac\pi6$$
Now you can plug $x$ in the formulas above and find out the general solutions for your problem.
A: 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 $k \in \mathbb{Z}$.
