General solution set of the trig function $\sin \theta = s$ The general solution set of $\sin \theta = \frac{1}{2}$ is:
$$\theta = \frac{\pi}{6}+2n\pi$$ and
$$\theta = \frac{5\pi}{6}+2n\pi$$
The book states that the two sets can be combined like so:
$$\theta = (-1)^n \frac{\pi}{6}+n\pi$$
It is non-obvious to me how this set includes the above two. Say, I want a solution from the PV (first) set, so that I take $n=1$. The output of general solution is $-\frac{\pi}{6}+\pi \equiv \frac{5\pi}{6}$, but should be $13\pi/6$ or $17\pi/6$ ?.
Then the book goes on to state that generally, if $\sin \theta = s$, then the general solution set is of the form $$\theta = (-1)^nPV+n\pi$$ Where PV is the principal  value, i.e. the unique solution that occurs in the domain $[-\pi/2,\pi/2]$, and of course $|s| \leq 1$.
Can someone prove in a simple way, that the general solution is indeed that and explain why I am not obtaining the correct solution values when using the general solution set formula.
 A: For your first question, here's the way you should think about it. For even $n$, we can write $n = 2m$ for some other positive integer $m$. Then we have:
$$ (-1)^n \frac{\pi}{6} + n \pi = \frac{\pi}{6} + 2m \pi $$
Note that this is the first set of solutions you have listed.
On the other hand, when $n$ is odd, we may write $n = 2m + 1$ for some integer $m \geq 0$. Then we have:
$$ (-1)^n \frac{\pi}{6} + n \pi = \frac{-\pi}{6} + (2m + 1)\pi $$
This is the second set of solutions you have listed. This proves your first claim.
For the second, you can follow this same argument. Here, the unit circle is your best friend. You want to picture all of the possible coterminal angles to $PV$; this just provides a short description of them all. It may also help you to remember that $\sin(\theta) = \sin(\pi - \theta)$.
A: since functions $y=\sin { \theta  } $ and $y=\frac { 1 }{ 2 } $ intersect in two points,equation have two roots $$ { x }_{ 1 }=\arcsin { \frac { 1 }{ 2 } +2n\pi  } =\frac { \pi  }{ 6 } +2n\pi \quad \quad \quad \quad \quad \left( 1 \right) \\ { x }_{ 2 }=\pi -\arcsin { \frac { 1 }{ 2 } +2n\pi  } =-\frac { \pi  }{ 6 } +\left( 2n+1 \right) \pi \quad \quad \quad \quad \quad \quad \left( 2 \right) \\ $$
when n is even then (1) is solutuin,when is odd then (2) is solutuion

